(a) To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{1}^{\frac{3}{2}} y^2 \, dx\)

Substitute \(y = \frac{4}{(2x-1)^2}\):

\(V = \pi \int_{1}^{\frac{3}{2}} \left( \frac{4}{(2x-1)^2} \right)^2 \, dx = \pi \int_{1}^{\frac{3}{2}} \frac{16}{(2x-1)^4} \, dx\)

Integrate:

\(\pi \left[ -\frac{16}{3 \times 2x(2x-1)^3} \right]_{1}^{\frac{3}{2}}\)

Calculate the definite integral:

\(\pi \left( \frac{112}{48} \right) = \frac{11}{6} \pi\)

(b) Differentiate \(y = \frac{4}{(2x-1)^2}\) to find the gradient at \(B\):

\(\frac{dy}{dx} = -8(2x-1)^{-3} \times 2\)

At \(B\), the gradient is \(-2\).

Equation of the tangent at \(B\):

\(y - 1 = -2\left(x - \frac{3}{2}\right)\)

Equation of the normal at \(B\):

\(y - 1 = \frac{1}{2}\left(x - \frac{3}{2}\right)\)

The tangent crosses the x-axis at 2, and the normal crosses the x-axis at \(\frac{1}{2}\).

Calculate the area of the triangle:

\(\text{Area} = \frac{1}{2} \times \left(2 - \frac{1}{2}\right) \times 1 = \frac{5}{4}\)

(i) Differentiate \(y = (x-1)^{-2} + 2\) to find the gradient of the tangent:

\(\frac{dy}{dx} = -2(x-1)^{-3}\).

At \(x = 2\), \(\frac{dy}{dx} = -2\), so the gradient of the normal is \(\frac{1}{2}\).

The equation of the normal through \((2, 3)\) is:

\(y - 3 = \frac{1}{2}(x - 2)\).

Simplifying gives \(y = \frac{1}{2}x + 2\).

(ii) The volume of revolution is given by:

\(\pi \int_{1}^{3} y_1^2 \, dx + \pi \int_{1}^{3} y_2^2 \, dx\).

For \(y_1 = \frac{1}{2}x + 2\), \(y_1^2 = \left(\frac{1}{2}x + 2\right)^2\).

For \(y_2 = (x-1)^{-2} + 2\), \(y_2^2 = \left((x-1)^{-2} + 2\right)^2\).

Integrate and evaluate:

\(\pi \left[ \frac{1}{3} \left(\frac{1}{2}x + 2\right)^3 \right]_{1}^{3} + \pi \left[ \frac{1}{12}x^2 + x + 4x \right]_{1}^{3}\).

Calculate the definite integrals and sum the volumes:

\(\frac{7}{12} \pi + \frac{6}{7} \pi = \frac{111}{8} \pi\) or \(13.9\pi\) or 43.6.

(i) To express \(y = x^2 + 4x + 3\) in the form \(y = (x + a)^2 + b\), complete the square:

\(y = (x + 2)^2 - 1\).

Hence, \(a = 2\) and \(b = -1\).

To express \(x\) in terms of \(y\):

\(y = (x + 2)^2 - 1\)

\(y + 1 = (x + 2)^2\)

\(x + 2 = \pm \sqrt{y + 1}\)

Since \(x \geq -2\), we take the positive root:

\(x = -2 + \sqrt{y + 1}\).

(ii) To find the volume when the shaded region is rotated about the y-axis, use the formula for the volume of revolution:

\(V = \pi \int_{-1}^{3} x^2 \, dy\)

Substitute \(x = -2 + \sqrt{y + 1}\):

\(x^2 = 4 + (y + 1) - 4(y + 1)^{1/2}\)

\(V = \pi \int_{-1}^{3} \left[ 5y + \frac{y^2}{2} - \frac{4(y + 1)^{3/2}}{3/2} \right] \, dy\)

Integrate and apply limits:

\(V = \pi \left[ 15 + \frac{9}{2} - \frac{64}{3} - \left(-5 + \frac{1}{2}\right) \right]\)

\(V = \frac{8\pi}{3}\) or 8.38

(i) To find the volume of the solid obtained by rotating the region about the x-axis, we use the formula:

\(V = \pi \int_0^3 (x^3 + x^2) \, dx\)

Integrating, we have:

\(\pi \left[ \frac{x^4}{4} + \frac{x^3}{3} \right]_0^3\)

\(= \pi \left[ \frac{81}{4} + 9 - 0 \right]\)

\(= \frac{117\pi}{4}\)

(ii) First, find the derivative \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{1}{2} (x^3 + x^2)^{-1/2} \times (3x^2 + 2x)\)

At \(x = 3\), \(y = 6\).

The gradient \(m\) at \(x = 3\) is:

\(m = \frac{1}{2} \times \frac{1}{6} \times 33 = \frac{11}{4}\)

The equation of the normal is:

\(y - 6 = -\frac{4}{11}(x - 3)\)

When \(x = 0\), \(y =7 \frac{1}{11}\).

(i) To find the volume of the solid of revolution, use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Substitute \(y = 2(3x - 1)^{-\frac{1}{3}}\):

\(V = \pi \int_{\frac{2}{3}}^{3} \left(2(3x - 1)^{-\frac{1}{3}}\right)^2 \, dx\)

\(V = 4\pi \int_{\frac{2}{3}}^{3} (3x - 1)^{-\frac{2}{3}} \, dx\)

Integrate:

\(V = 4\pi \left[ \frac{(3x - 1)^{\frac{1}{3}}}{\frac{1}{3}} \right]_{\frac{2}{3}}^{3}\)

\(V = 4\pi \left[ 3(3x - 1)^{\frac{1}{3}} \right]_{\frac{2}{3}}^{3}\)

Apply the limits:

\(V = 4\pi [2 - 1]\)

\(V = 4\pi\)

Thus, the volume is \(4\pi\) or approximately 12.6.

(ii) Differentiate \(y = 2(3x - 1)^{-\frac{1}{3}}\) to find the gradient:

\(\frac{dy}{dx} = (-\frac{2}{3})(3x - 1)^{-\frac{4}{3}} \times 3\)

\(\frac{dy}{dx} = -2(3x - 1)^{-\frac{4}{3}}\)

At \(x = \frac{2}{3}\), \(y = 2\), and \(\frac{dy}{dx} = -2\).

The gradient of the normal is \(\frac{1}{2}\).

The equation of the normal is:

\(y - 2 = \frac{1}{2}(x - \frac{2}{3})\)

\(y = \frac{1}{2}x + \frac{5}{3}\)

(i) Differentiate \(y = (x+1)^2 + (x+1)^{-1}\) to find \(\frac{dy}{dx} = 2(x+1) - (x+1)^{-2}\).

Set \(\frac{dy}{dx} = 0\) to find the minimum point: \(2(x+1)^3 = 1\).

Differentiate again to find \(\frac{d^2y}{dx^2} = 2 + 2(x+1)^{-3}\).

Substitute \((x+1)^{-3} = 2\) to find \(\frac{d^2y}{dx^2} = 6\).

(ii) The volume of the solid of revolution is given by \(\pi \int y^2 \, dx\).

Calculate \(y^2 = (x+1)^4 + (x+1)^{-2} + 2(x+1)^2\).

Integrate \(y^2\) from 0 to 1: \(\pi \left[ \frac{(x+1)^5}{5} + (x+1)^{-1} + \frac{2(x+1)^2}{2} \right]_0^1\).

Evaluate the definite integral: \(\pi \left[ \frac{32}{5} - \frac{1}{2} + 4 - \left( \frac{1}{5} + 1 \right) \right]\).

The volume is \(9.7\pi\) or \(30.5\).

(i) To find the points \(P\) and \(Q\), set \(y = \frac{x}{2} + \frac{6}{x} = 4\). Solving gives \(x = 2\) and \(x = 6\). The points are \(P(2, 4)\) and \(Q(6, 4)\).

The derivative of \(y\) is \(\frac{dy}{dx} = \frac{1}{2} - \frac{6}{x^2}\).

At \(x = 2\), the slope \(m = \frac{1}{2} - \frac{6}{4} = -1\). The tangent line is \(y - 4 = -1(x - 2)\) or \(y = -x + 6\).

At \(x = 6\), the slope \(m = \frac{1}{2} - \frac{6}{36} = \frac{1}{3}\). The tangent line is \(y - 4 = \frac{1}{3}(x - 6)\) or \(y = \frac{1}{3}x + 2\).

Solving \(y = -x + 6\) and \(y = \frac{1}{3}x + 2\) gives \(x = 3\), \(y = 3\). The tangents meet at \((3, 3)\), which lies on \(y = x\).

(ii) The volume of revolution is given by \(V = \pi \int_2^6 \left(4^2 - \left(\frac{x}{2} + \frac{6}{x}\right)^2\right) \, dx\).

Integrate \(\pi \int_2^6 \left(16 - \left(\frac{x}{2} + \frac{6}{x}\right)^2\right) \, dx\).

\(= \pi \int_2^6 \left(16 - \left(\frac{x^2}{4} + 6 + \frac{36}{x^2}\right)\right) \, dx\)

\(= \pi \int_2^6 \left(\frac{-x^2}{4} + 10 - \frac{36}{x^2}\right) \, dx\)

Integrate to get \(\left[-\frac{x^3}{12} + 10x + \frac{36}{x}\right]_2^6\).

Calculate: \(\left(-\frac{6^3}{12} + 10 \times 6 + \frac{36}{6}\right) - \left(-\frac{2^3}{12} + 10 \times 2 + \frac{36}{2}\right)\)

\(= \left(-18 + 60 + 6\right) - \left(-\frac{8}{12} + 20 + 18\right)\)

\(= 48 - 37 = 11\)

Volume \(= \pi \times 11 = \frac{10}{3} \pi\) or approximately \(33.5 \pi\).

(i) The area of the shaded region is given by the integral:

\(\frac{1}{2} \int_0^1 (x^4 - 1) \, dx\)

Integrating, we have:

\(\frac{1}{2} \left[ \frac{x^5}{5} - x \right]_0^1\)

\(= \frac{1}{2} \left( \frac{1}{5} - 1 \right) = -\frac{2}{5}\)

The area is \(\frac{2}{5}\).

(ii) The volume when rotated about the x-axis is:

\(\pi \int_0^1 y^2 \, dx = \frac{\pi}{4} \int_0^1 (x^8 - 2x^4 + 1) \, dx\)

Integrating, we have:

\(\frac{\pi}{4} \left[ \frac{x^9}{9} - \frac{2x^5}{5} + x \right]_0^1\)

\(= \frac{\pi}{4} \left( \frac{1}{9} - \frac{2}{5} + 1 \right) = \frac{8\pi}{45}\)

The volume is \(\frac{8\pi}{45}\) or 0.559.

(iii) The volume when rotated about the y-axis is:

\(\pi \int_{-\frac{1}{2}}^0 x^2 \, dy = (\pi) \int_{-\frac{1}{2}}^0 (2y+1)^{1/2} \, dy\)

Integrating, we have:

\((\pi) \left[ \frac{(2y+1)^{3/2}}{3/2} \right]_{-\frac{1}{2}}^0\)

\(= (\pi) \left[ \frac{1}{3} - 0 \right] = \frac{\pi}{3}\)

The volume is \(\frac{\pi}{3}\) or 1.05.

(i) To find the volume of revolution, we use the formula \(V = \pi \int (y + 1) \, dy\) from \(y = 0\) to \(y = h\).

Integrating, we have:

\(V = \pi \left[ \frac{y^2}{2} + y \right]_0^h\)

\(= \pi \left( \frac{h^2}{2} + h \right)\)

Thus, \(V = \pi \left( \frac{1}{2}h^2 + h \right)\).

(ii) To find the area of the shaded region when \(h = 3\), we calculate:

\(\int (y + 1)^{1/2} \, dy\) from \(y = 0\) to \(y = 3\).

Alternatively, \(6 - \int (x^2 - 1) \, dx\) from \(x = 1\) to \(x = 2\).

Integrating, we have:

\(\left[ \frac{2}{3}(y + 1)^{3/2} \right]_0^3\)

\(= \frac{2}{3} \left[ 8 - 1 \right]\)

\(= \frac{14}{3}\)

The volume of the solid of revolution is given by the formula:

\(V = \pi \int (f(x)^2 - g(x)^2) \, dx\)

where \(f(x) = 5 - x\) and \(g(x) = \frac{4}{x}\).

Calculate the volume by integrating from \(x = 1\) to \(x = 4\):

\(V = \pi \left( \int_1^4 (5-x)^2 \, dx - \int_1^4 \frac{16}{x^2} \, dx \right)\)

First, integrate \((5-x)^2\):

\(\int (5-x)^2 \, dx = \frac{(5-x)^3}{3}\)

Evaluate from 1 to 4:

\(\left[ \frac{(5-x)^3}{3} \right]_1^4 = \frac{(5-4)^3}{3} - \frac{(5-1)^3}{3} = \frac{1}{3} - \frac{64}{3} = -\frac{63}{3} = -21\)

Next, integrate \(\frac{16}{x^2}\):

\(\int \frac{16}{x^2} \, dx = -\frac{16}{x}\)

Evaluate from 1 to 4:

\(\left[ -\frac{16}{x} \right]_1^4 = -\frac{16}{4} + \frac{16}{1} = -4 + 16 = 12\)

Subtract the two results:

\(V = \pi (-21 - 12) = \pi (-33) = 33\pi\)

Thus, the volume is \(9\pi\) or approximately 28.3.

(i) To find the equation of the normal, first find the derivative \(\frac{dy}{dx}\) of the curve \(y = \frac{4}{5 - 3x}\).

\(\frac{dy}{dx} = \frac{-4}{(5 - 3x)^2} \times (-3) = \frac{12}{(5 - 3x)^2}\).

At \(x = 1\), \(\frac{dy}{dx} = \frac{12}{(5 - 3 \times 1)^2} = \frac{12}{4} = 3\).

The gradient of the tangent is 3, so the gradient of the normal is \(-\frac{1}{3}\).

The point on the curve at \(x = 1\) is \((1, 2)\).

The equation of the normal is \(y - 2 = -\frac{1}{3}(x - 1)\).

Simplifying gives \(y = -\frac{1}{3}x + \frac{7}{3}\).

(ii) The volume of the solid of revolution is given by \(V = \pi \int_{0}^{1} y^2 \, dx\).

\(V = \pi \int_{0}^{1} \left( \frac{4}{5 - 3x} \right)^2 \, dx = \pi \int_{0}^{1} \frac{16}{(5 - 3x)^2} \, dx\).

Let \(u = 5 - 3x\), then \(du = -3 \, dx\) or \(dx = -\frac{1}{3} \, du\).

Change the limits: when \(x = 0, u = 5\); when \(x = 1, u = 2\).

\(V = \pi \int_{5}^{2} \frac{16}{u^2} \left(-\frac{1}{3}\right) \, du\).

\(V = -\frac{16\pi}{3} \int_{5}^{2} \frac{1}{u^2} \, du\).

\(V = -\frac{16\pi}{3} \left[ -\frac{1}{u} \right]_{5}^{2}\).

\(V = \frac{16\pi}{3} \left( \frac{1}{2} - \frac{1}{5} \right)\).

\(V = \frac{16\pi}{3} \times \frac{3}{10} = \frac{8\pi}{5}\).

(a) To find the equation of the line \(AB\), first calculate the gradient:

\(\text{Gradient of } AB = \frac{2 - (-1)}{5 - 2} = 1\)

Using point \(A(5, 2)\), the equation of the line is:

\(y - 2 = 1(x - 5)\)

Simplifying gives:

\(y = x - 3\)

(b) To find the volume of revolution, calculate the integral of the area between the curve \(x = y^2 + 1\) and the line \(y = x - 3\) rotated about the \(y\)-axis:

For the curve:

\(\pi \int (y^2 + 1)^2 \, dy = \pi \int (y^4 + 2y^2 + 1) \, dy\)

Integrating gives:

\(\pi \left( \frac{y^5}{5} + \frac{2y^3}{3} + y \right)\)

For the line:

\(\pi \int (y + 3)^2 \, dy = \pi \int (y^2 + 6y + 9) \, dy\)

Integrating gives:

\(\pi \left( \frac{y^3}{3} + 3y^2 + 9y \right)\)

Apply limits from \(y = -1\) to \(y = 2\):

\(\pi \left( \frac{8}{3} + 12 + 18 \right) - \pi \left( -\frac{1}{3} + 3 - 9 \right)\)

Calculating gives:

\(\pi \left( 39 - 15 \frac{3}{5} \right)\)

Volume = \(\frac{117}{5} \pi\) or approximately 73.5

(i) Differentiate \(y = (kx - 3)^{-1} + (kx - 3)\) to find \(\frac{dy}{dx} = -k(kx - 3)^{-2} + k\). Set \(\frac{dy}{dx} = 0\) to find stationary points:

\(-k(kx - 3)^{-2} + k = 0\)

\((kx - 3)^{2} = 1\)

\(k^{2}x^{2} - 6kx + 8k = 0\)

Solving gives \(x = \frac{2}{k}\) or \(x = \frac{4}{k}\).

To determine the nature, find \(\frac{d^{2}y}{dx^{2}} = 2k^{2}(kx - 3)^{-3}\).

When \(x = \frac{2}{k}\), \(\frac{d^{2}y}{dx^{2}} = (-2k^{2}) < 0\), so it is a maximum.

When \(x = \frac{4}{k}\), \(\frac{d^{2}y}{dx^{2}} = (2k^{2}) > 0\), so it is a minimum.

(ii) For \(k = 1\), the curve is \(y = (x - 3)^{-1} + (x - 3)\). The volume \(V\) is given by:

\(V = \pi \int_{0}^{2} [(x - 3)^{-1} + (x - 3)]^{2} \, dx\)

\(= \pi \int_{0}^{2} [(x - 3)^{-2} + (x - 3)^{2} + 2] \, dx\)

\(= \pi \left[ -(x - 3)^{-1} + \frac{(x - 3)^{3}}{3} + 2x \right]_{0}^{2}\)

\(= \pi \left[ -\frac{1}{3} + 4 - \left(-\frac{1}{3} - 9 + 0\right) \right]\)

\(= \frac{40\pi}{3}\) or 41.9

To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Given \(y = (x^3 + 1)^{\frac{1}{2}}\), we have:

\(y^2 = x^3 + 1\)

Thus, the volume is:

\(V = \pi \int_{0}^{2} (x^3 + 1) \, dx\)

Integrating, we get:

\(\pi \left[ \frac{x^4}{4} + x \right]_{0}^{2}\)

Evaluating the definite integral:

\(\pi \left( \left( \frac{2^4}{4} + 2 \right) - \left( \frac{0^4}{4} + 0 \right) \right)\)

\(= \pi \left( 4 + 2 \right)\)

\(= 6\pi\)

Thus, the volume is \(6\pi\) or approximately 18.8.

(i) To find \(\frac{dy}{dx}\), differentiate \(y = \frac{8}{x} + 2x\):

\(\frac{dy}{dx} = -8x^{-2} + 2\).

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\):

\(\frac{d^2y}{dx^2} = 16x^{-3}\).

To find \(\int y^2 \, dx\), integrate \(y^2 = \left(\frac{8}{x} + 2x\right)^2\):

\(\int y^2 \, dx = -64x^{-1} + 32x + \frac{4x^3}{3} + C\).

(ii) Set \(\frac{dy}{dx} = 0\) to find stationary points:

\(-8x^{-2} + 2 = 0 \Rightarrow x = \pm 2\).

For \(x > 0\), \(M(2, 8)\).

For \(x < 0\), the other turning point is \((-2, -8)\).

Check the nature using \(\frac{d^2y}{dx^2}\):

At \(x = -2\), \(\frac{d^2y}{dx^2} < 0\), so it is a maximum.

(iii) The volume of revolution is given by:

\(V = \pi \int_{1}^{2} y^2 \, dx\).

Using the integral from part (i),

\(V = \pi \left[ -64x^{-1} + 32x + \frac{4x^3}{3} \right]_{1}^{2}\).

Calculate the definite integral:

\(V = \frac{220\pi}{3}\).

To find the volume of the solid formed when the shaded region is rotated about the y-axis, we use the formula for the volume of revolution:

\(V = \pi \int_{a}^{b} x^2 \, dy\)

Given \(x = \frac{12}{y^2} - 2\), we have:

\(x^2 = \left( \frac{12}{y^2} - 2 \right)^2\)

Expanding, \(x^2 = \frac{144}{y^4} - \frac{48}{y^2} + 4\).

The volume is:

\(V = \pi \int_{1}^{2} \left( \frac{144}{y^4} - \frac{48}{y^2} + 4 \right) \, dy\)

Integrating term by term:

\(\int \frac{144}{y^4} \, dy = -\frac{144}{3y^3} = -\frac{48}{y^3}\)

\(\int \frac{48}{y^2} \, dy = -\frac{48}{y}\)

\(\int 4 \, dy = 4y\)

Thus,

\(V = \pi \left[ -\frac{48}{y^3} - \frac{48}{y} + 4y \right]_{1}^{2}\)

Evaluating from 1 to 2:

\(V = \pi \left( \left( -\frac{48}{8} - \frac{48}{2} + 8 \right) - \left( -48 - 48 + 4 \right) \right)\)

\(V = \pi \left( -6 - 24 + 8 + 92 \right)\)

\(V = \pi \times 22\)

Therefore, the volume is \(22\pi\).

(i) The curve is given by \(y = \sqrt{9 - 2x^2}\). The derivative is \(\frac{dy}{dx} = \frac{1}{2\sqrt{9 - 2x^2}} \times (-4x)\).

At \(P(2, 1)\), \(x = 2\), so \(m = -4\). The gradient of the normal is \(\frac{1}{4}\).

The equation of the normal at \(P\) is \(y - 1 = \frac{1}{4}(x - 2)\).

Solving for \(A\) and \(B\):

\(A(-2, 0)\) and \(B(0, \frac{1}{2})\).

The midpoint of \(AP\) is \((0, \frac{1}{2})\), which is \(B\).

(ii) The volume of the solid of revolution is given by:

\(\int x^2 \, dy = \int \left( \frac{9}{2} - \frac{y^2}{2} \right) \, dy\).

\(= \frac{9y}{2} - \frac{y^3}{6}\).

Using limits from 1 to 3:

Volume = \(\left[ \frac{9y}{2} - \frac{y^3}{6} \right]_1^3 = \frac{44}{3} \pi\).

(i) To show that \(PQ\) is a normal to the curve, we first find the derivative of the curve \(y = (1 + 4x)^{\frac{1}{2}}\).

The derivative is \(\frac{dy}{dx} = \frac{1}{2}(1 + 4x)^{-\frac{1}{2}} \times 4 = 2(1 + 4x)^{-\frac{1}{2}}\).

At \(x = 6\), \(\frac{dy}{dx} = \frac{2}{5}\).

The gradient of the normal at \(P\) is \(-\frac{1}{2}\).

The gradient of \(PQ\) is \(-\frac{5}{2}\), hence \(PQ\) is a normal, or \(m_1 m_2 = -1\).

(ii) To find the volume of revolution, we calculate the volume for the curve and the line separately.

Volume for the curve: \(V = \pi \int (1 + 4x) \, dx\).

Integrating \(y^2 = (1 + 4x)\), we get \(\pi \left[ x + 2x^2 \right]\) from 0 to 6.

\(= \pi [6 + 72 - 0] = 78\pi\).

Volume for the line: \(V = \frac{1}{3} \pi \times 5^2 \times 2 = \frac{50}{3} \pi\).

Total volume: \(78\pi + \frac{50}{3}\pi = 94\frac{2}{3}\pi\) or \(\frac{284\pi}{3}\).

(i) To find the volume of the solid of revolution, use the formula:

\(V = \pi \int_{1}^{2} y^2 \, dx\)

Substitute \(y = \frac{4}{2x-1}\):

\(V = \pi \int_{1}^{2} \left( \frac{4}{2x-1} \right)^2 \, dx = \pi \int_{1}^{2} \frac{16}{(2x-1)^2} \, dx\)

Integrate:

\(\int \frac{16}{(2x-1)^2} \, dx = -\frac{16}{2x-1}\)

Evaluate from 1 to 2:

\(V = \pi \left[ -\frac{16}{2x-1} \right]_{1}^{2} = \pi \left( -\frac{16}{3} + 16 \right) = \frac{16\pi}{3}\)

(ii) The slope of the tangent to the curve is given by the derivative:

\(\frac{dy}{dx} = \frac{-4}{(2x-1)^2}\)

The slope of the normal is \(m = -\frac{1}{2} \times (-2) = 2\).

Equating \(\frac{dy}{dx} = -2\):

\(\frac{-4}{(2x-1)^2} = -2\)

Solve for \(x\):

\(x = \frac{3}{2} \text{ or } x = -\frac{1}{2}\)

For \(x = \frac{3}{2}\), \(y = 2\), so \(2y = x + c\) gives \(4 = \frac{3}{2} + c\), \(c = \frac{5}{2}\).

For \(x = -\frac{1}{2}\), \(y = -2\), so \(2y = x + c\) gives \(-4 = -\frac{1}{2} + c\), \(c = -\frac{7}{2}\).

To find the volume of the solid of revolution when the region is rotated about the y-axis, we use the formula:

\(V = \pi \int x^2 \, dy\)

Given \(y = x^2 + 1\), we have \(x^2 = y - 1\).

Thus, the integral becomes:

\(V = \pi \int (y - 1) \, dy\)

We integrate with respect to \(y\) from 1 to 5:

\(V = \pi \left[ \frac{1}{2}y^2 - y \right]_1^5\)

Calculating the definite integral:

\(V = \pi \left( \frac{1}{2}(5)^2 - 5 - \left( \frac{1}{2}(1)^2 - 1 \right) \right)\)

\(V = \pi \left( \frac{25}{2} - 5 - \frac{1}{2} + 1 \right)\)

\(V = \pi \left( 12.5 - 5 - 0.5 + 1 \right)\)

\(V = \pi \times 8\)

\(V = 8\pi\)

Thus, the volume is \(8\pi\) or approximately 25.1.

To find the volume of revolution, we use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Given \(y = \sqrt{x^4 + 4x + 4}\), we have:

\(y^2 = x^4 + 4x + 4\)

Integrate \(y^2\) from \(x = -1\) to \(x = 0\):

\(\int (x^4 + 4x + 4) \, dx = \left[ \frac{x^5}{5} + 2x^2 + 4x \right]\)

Apply the limits:

\(\pi \left[ \left( \frac{0^5}{5} + 2(0)^2 + 4(0) \right) - \left( \frac{(-1)^5}{5} + 2(-1)^2 + 4(-1) \right) \right]\)

\(= \pi \left[ 0 - \left( -\frac{1}{5} + 2 - 4 \right) \right]\)

\(= \pi \left[ 0 - \left( -\frac{1}{5} - 2 \right) \right]\)

\(= \pi \left[ \frac{1}{5} + 2 \right]\)

\(= \pi \left[ \frac{11}{5} \right]\)

Thus, the volume of revolution is \(\frac{11\pi}{5}\).

To find the volume of the solid formed by rotating the region under the curve \(y = \frac{8}{x} + 2x\) from \(x = 2\) to \(x = 5\) about the \(x\)-axis, we use the formula for the volume of revolution:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Substitute \(y = \frac{8}{x} + 2x\) into the formula:

\(V = \pi \int_{2}^{5} \left( \frac{8}{x} + 2x \right)^2 \, dx\)

Expand \(\left( \frac{8}{x} + 2x \right)^2\):

\(\left( \frac{8}{x} + 2x \right)^2 = \frac{64}{x^2} + 4x^2 + 32\)

Integrate each term separately:

\(\int \frac{64}{x^2} \, dx = -\frac{64}{x}\)

\(\int 4x^2 \, dx = \frac{4x^3}{3}\)

\(\int 32 \, dx = 32x\)

Combine the integrals:

\(V = \pi \left[ -\frac{64}{x} + \frac{4x^3}{3} + 32x \right]_{2}^{5}\)

Evaluate from \(x = 2\) to \(x = 5\):

\(V = \pi \left( \left( -\frac{64}{5} + \frac{4(5)^3}{3} + 32(5) \right) - \left( -\frac{64}{2} + \frac{4(2)^3}{3} + 32(2) \right) \right)\)

Calculate the values:

\(V = \pi \left( -12.8 + \frac{500}{3} + 160 - (-32 + \frac{32}{3} + 64) \right)\)

\(V = \pi \left( 271.2 \right)\)

Thus, the volume is \(271.2\pi\) or approximately 852.

(a) To find the coordinates of \(A\), solve the system of equations given by the circle \(x^2 + y^2 = 2\) and the line \(y = 2x - 1\). Substitute \(y = 2x - 1\) into the circle equation:

\(x^2 + (2x - 1)^2 = 2\)

\(x^2 + 4x^2 - 4x + 1 = 2\)

\(5x^2 - 4x - 1 = 0\)

Solving this quadratic equation, we find \(x = 1\). Substituting back, \(y = 2(1) - 1 = 1\). Thus, \(A = (1, 1)\).

(b) The volume of revolution is given by \(\pi \int (2 - x^2) \, dx\) from \(x = 1\) to \(x = \sqrt{2}\). The integral is:

\(\pi \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt{2}}\)

\(\pi \left[ 2\sqrt{2} - \frac{(\sqrt{2})^3}{3} - (2 - \frac{1}{3}) \right]\)

\(\pi \left[ 2\sqrt{2} - \frac{2\sqrt{2}}{3} - \frac{5}{3} \right]\)

\(\frac{\pi}{3}(4\sqrt{2} - 5)\)

(c) The perimeter of the shaded region consists of the line segment \(AD\) and the arc length. The arc length is \(\frac{1}{8}(2\pi\sqrt{2})\) or \(\frac{\pi\sqrt{2}}{4}\). Therefore, the perimeter is:

\(\sqrt{2} + \frac{\pi\sqrt{2}}{4}\)

To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} y^2 \, dx\)

Given \(y = \frac{6}{2x-3}\), we have:

\(y^2 = \left(\frac{6}{2x-3}\right)^2 = \frac{36}{(2x-3)^2}\)

The integral becomes:

\(V = \pi \int_{2}^{3} \frac{36}{(2x-3)^2} \, dx\)

Using the substitution \(u = 2x-3\), \(du = 2 \, dx\), \(dx = \frac{du}{2}\), the limits change from \(x = 2\) to \(u = 1\) and \(x = 3\) to \(u = 3\).

The integral becomes:

\(V = \pi \int_{1}^{3} \frac{36}{u^2} \cdot \frac{1}{2} \, du = 18\pi \int_{1}^{3} u^{-2} \, du\)

Integrating, we get:

\(18\pi \left[ -u^{-1} \right]_{1}^{3} = 18\pi \left( -\frac{1}{3} + 1 \right) = 18\pi \left( \frac{2}{3} \right) = 12\pi\)

Thus, the volume is \(12\pi\).

(i) Start with the equation \(y = \frac{2}{\sqrt{x+1}}\).

Square both sides: \(y^2 = \frac{4}{x+1}\).

Rearrange to find \(x\): \(x+1 = \frac{4}{y^2}\).

Thus, \(x = \frac{4}{y^2} - 1\).

(ii) Integrate \(\int \left( \frac{4}{y^2} - 1 \right) \, dy\).

The integral is \(\left[ -\frac{4}{y} - y \right]\).

Apply limits from 1 to 2: \(\left[ -\frac{4}{2} - 2 \right] - \left[ -\frac{4}{1} - 1 \right]\).

Calculate: \(\left[ -2 - 2 \right] - \left[ -4 - 1 \right] = -4 + 5 = 1\).

(iii) Volume of revolution: \(\pi \int x^2 \, dy = \pi \int \left( \frac{16}{y^4} - \frac{8}{y^2} + 1 \right) \, dy\).

The integral is \(\pi \left[ -\frac{16}{3y^3} + \frac{8}{y} + y \right]\).

Apply limits from 1 to 2: \(\pi \left[ -\frac{16}{24} + 4 + 2 \right] - \pi \left[ -\frac{16}{3} + 8 + 1 \right]\).

Calculate: \(\pi \left[ -\frac{2}{3} + 6 \right] - \pi \left[ -\frac{16}{3} + 9 \right] = \pi \left[ \frac{16}{3} \right] = \frac{5\pi}{3}\).

(i) To find the area of the shaded region, we calculate the integral of the difference between the curve and the line from \(x = -1\) to \(x = 0\):

\(\int_{-1}^{0} \left( (x+1)^{\frac{1}{2}} - (x+1) \right) \, dx\)

Calculate the integrals separately:

\(\int (x+1)^{\frac{1}{2}} \, dx = \frac{2}{3}(x+1)^{\frac{3}{2}}\)

\(\int (x+1) \, dx = \frac{1}{2}x^2 + x\)

Apply the limits:

\(\frac{2}{3}(0+1)^{\frac{3}{2}} - \left( \frac{1}{2}(0)^2 + 0 \right) - \left( \frac{2}{3}(-1+1)^{\frac{3}{2}} - \left( \frac{1}{2}(-1)^2 - 1 \right) \right)\)

\(= \frac{2}{3} - 0 - (0 - \frac{1}{2}) = \frac{1}{6}\)

(ii) To find the volume of the solid obtained by rotating the shaded region about the y-axis, use the method of cylindrical shells:

\(V = \pi \int_{0}^{1} \left( (y^2 - 1)^2 - (y-1)^2 \right) \, dy\)

Calculate the integrals:

\(\int (y^4 - 2y^2 + 1) \, dy = \frac{y^5}{5} - \frac{2y^3}{3} + y\)

Apply the limits:

\(\pi \left[ \frac{1}{5} - \frac{2}{3} + 1 \right] = \frac{8}{15} \pi\)

Alternatively, using the volume of a cone:

\(V = \frac{1}{3} \pi (x_1^2 - x_2^2)\)

\(= \frac{1}{5} \pi\)

(i) To find \(\frac{dy}{dx}\), use the chain rule. Let \(u = 8x - x^2\), then \(y = \sqrt{u} = u^{\frac{1}{2}}\).

\(\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}}\)

\(\frac{du}{dx} = 8 - 2x\)

Thus, \(\frac{dy}{dx} = \frac{1}{2}(8x-x^2)^{-\frac{1}{2}} \times (8-2x)\).

For the stationary point, set \(\frac{dy}{dx} = 0\):

\((8-2x) = 0 \Rightarrow x = 4\).

Substitute \(x = 4\) into \(y = \sqrt{(8x - x^2)}\):

\(y = \sqrt{(8 \times 4 - 4^2)} = \sqrt{16} = 4\).

Thus, the stationary point is (4, 4).

(ii) To find the volume, use the formula for the volume of revolution:

\(V = \pi \int (8x - x^2) \, dx\) from \(x = 0\) to \(x = 8\).

\(V = \pi \left[ \frac{4x^2}{2} - \frac{x^3}{3} \right]_0^8\)

\(V = \pi \left[ \frac{4 \times 8^2}{2} - \frac{8^3}{3} \right]\)

\(V = \pi \left[ 128 - \frac{512}{3} \right]\)

\(V = \pi \left[ \frac{384}{3} \right]\)

\(V = \frac{256\pi}{3}\).

(i) To find the coordinates of \(B\), set \(x = 0\) in the equation \(y = \sqrt{1 + 2x}\):

\(y = \sqrt{1 + 2(0)} = \sqrt{1} = 1\)

So, \(B = (0, 1)\).

For \(C\), given \(y = 3\):

\(3 = \sqrt{1 + 2x}\)

\(9 = 1 + 2x\)

\(2x = 8\)

\(x = 4\)

So, \(C = (4, 3)\).

(ii) The derivative \(\frac{dy}{dx}\) is:

\(\frac{dy}{dx} = \frac{1}{2} \times 2(1 + 2x)^{-\frac{1}{2}} = \frac{1}{\sqrt{1 + 2x}}\)

At \(C(4, 3)\):

\(\frac{dy}{dx} = \frac{1}{\sqrt{9}} = \frac{1}{3}\)

The gradient of the normal is \(-3\).

Equation of the normal:

\(y - 3 = -3(x - 4)\)

or \(y = -3x + 15\).

(iii) To find the volume, use the formula for the volume of revolution:

\(y^2 = 1 + 2x \Rightarrow x = \frac{1}{2}(y^2 - 1)\)

Volume \(V = \pi \int_0^1 x^2 \, dy\)

\(V = \pi \int_0^1 \left( \frac{1}{2}(y^2 - 1) \right)^2 \, dy\)

\(V = \pi \times \frac{1}{4} \int_0^1 (y^4 - 2y^2 + 1) \, dy\)

\(V = \pi \times \frac{1}{4} \left[ \frac{y^5}{5} - \frac{2y^3}{3} + y \right]_0^1\)

\(V = \pi \times \frac{1}{4} \left[ \frac{1}{5} - \frac{2}{3} + 1 \right]\)

\(V = \frac{2}{15} \pi\)

(i) To find the coordinates of \(A\), set \(y = 0\) in the equation \(y = 4\sqrt{x} - x\):

\(4\sqrt{x} - x = 0\)

\(4\sqrt{x} = x\)

\(4 = \sqrt{x}\)

\(x = 16\)

Thus, \(A(16, 0)\).

To find the coordinates of \(M\), find the derivative \(\frac{dy}{dx} = 2x^{-\frac{1}{2}} - 1\) and set it to zero:

\(2x^{-\frac{1}{2}} - 1 = 0\)

\(2\sqrt{x} = 1\)

\(\sqrt{x} = \frac{1}{2}\)

\(x = 4\)

Substitute \(x = 4\) back into the original equation:

\(y = 4\sqrt{4} - 4 = 4\)

Thus, \(M(4, 4)\).

(ii) The volume \(V\) of the solid of revolution is given by:

\(V = \pi \int y^2 \, dx\)

\(V = \pi \int (4\sqrt{x} - x)^2 \, dx\)

\(V = \pi \int (16x + x^2 - 8x^{\frac{3}{2}}) \, dx\)

Integrate each term:

\(\pi \left[ 8x^2 + \frac{x^3}{3} - 8\frac{x^{\frac{5}{2}}}{\frac{5}{2}} \right]\)

Evaluate from 0 to 16:

\(V = \pi \left[ 8(16)^2 + \frac{(16)^3}{3} - 8\frac{(16)^{\frac{5}{2}}}{\frac{5}{2}} \right]\)

\(V = 136.5\pi\) (or \(137\pi\))

(i) The curve \(y = (x - 2)^2\) is a parabola opening upwards with vertex at \((2, 0)\). It touches the positive \(x\)-axis at \(x = 2\).

(ii) To find the volume of the solid of revolution, use the formula for the volume of a solid of revolution about the \(x\)-axis: \(V = \pi \int (y)^2 \, dx\).

Here, \(y = (x - 2)^2\), so \(y^2 = (x - 2)^4\).

Calculate the integral: \(V = \pi \int_0^2 (x - 2)^4 \, dx\).

Expand \((x - 2)^4\) and integrate: \(\pi \left[ \frac{(x - 2)^5}{5} \right]_0^2\).

Evaluate the integral: \(\pi \left[ \frac{(2 - 2)^5}{5} - \frac{(-2)^5}{5} \right] = \pi \left[ 0 - \frac{-32}{5} \right] = \frac{32\pi}{5}\).

Thus, the volume is \(\frac{32\pi}{5}\) or \(6.4\pi\).

(i) The curve cuts the y-axis at \(A = (0, 1)\) and the line \(x = 5\) at \(B = (5, \frac{1}{2})\).

The slope of line \(AB\) is \(\frac{\frac{1}{2} - 1}{5 - 0} = -\frac{1}{10}\).

Using the point-slope form: \(y - 1 = -\frac{1}{10}(x - 0)\).

Simplifying gives: \(y = -\frac{1}{10}x + 1\).

(ii) To find the volume, calculate the volume of revolution for both the curve and the line, then subtract.

For the curve: \(V_{\text{curve}} = \pi \int_{0}^{5} \left( \frac{1}{(3x+1)^{\frac{1}{4}}} \right)^2 \, dx\).

\(= \pi \int_{0}^{5} (3x+1)^{-\frac{1}{2}} \, dx\).

Integrating gives: \(\frac{2\pi}{3} [(3x+1)^{\frac{1}{2}}]_{0}^{5}\).

\(= \frac{2\pi}{3} [4 - 1] = 2\pi\).

For the line: \(V_{\text{line}} = \pi \int_{0}^{5} \left( -\frac{1}{10}x + 1 \right)^2 \, dx\).

\(= \pi \int_{0}^{5} \left( \frac{1}{100}x^2 - \frac{1}{5}x + 1 \right) \, dx\).

Integrating gives: \(\pi \left[ \frac{1}{300}x^3 - \frac{1}{10}x^2 + x \right]_{0}^{5}\).

\(= \pi \left[ \frac{125}{300} - \frac{25}{10} + 5 \right] = \frac{35\pi}{12}\).

The volume of the shaded region is \(V_{\text{line}} - V_{\text{curve}} = \frac{35\pi}{12} - 2\pi = \frac{11\pi}{12}\).

(a) To find the coordinates of \(A\), substitute \(x = 0\) into the circle equation \((x-2)^2 + y^2 = 8\):

\((-2)^2 + y^2 = 8\)

\(4 + y^2 = 8\)

\(y^2 = 4\)

\(y = 2\) (since \(A\) is on the positive \(y\)-axis)

Thus, \(A = (0, 2)\).

For \(B\), since \(AB\) is parallel to the \(x\)-axis, \(y = 2\). Substitute \(y = 2\) into the circle equation:

\((x-2)^2 + 4 = 8\)

\((x-2)^2 = 4\)

\(x-2 = \pm 2\)

\(x = 4\) (since \(B\) is on the right side of the circle)

Thus, \(B = (4, 2)\).

(b) The volume of revolution is found by integrating the area of the segment rotated about the \(x\)-axis:

Volume = \(\pi \int_0^4 (8 - (x-2)^2) \, dx\)

\(= \pi \left[ 8x - \frac{(x-2)^3}{3} \right]_0^4\)

\(= \pi \left[ 32 - \frac{16}{3} \right]\)

Volume of cylinder = \(\pi \times 2^2 \times 4 = 16\pi\)

Total volume = \(26\frac{2}{3}\pi - 16\pi = 10\frac{2}{3}\pi\)

The volume \(V\) of the solid of revolution is given by the formula:

\(V = \pi \int_0^1 y^2 \, dx\)

Substitute \(y = \frac{9}{2-x}\) into the formula:

\(V = \pi \int_0^1 \left(\frac{9}{2-x}\right)^2 \, dx\)

\(V = \pi \int_0^1 \frac{81}{(2-x)^2} \, dx\)

Integrate \(\frac{81}{(2-x)^2}\):

\(\int \frac{81}{(2-x)^2} \, dx = -81 \left(2-x\right)^{-1} \times (-1)\)

\(= 81 \left(2-x\right)^{-1}\)

Evaluate from 0 to 1:

\(V = \pi \left[ 81 \left(2-1\right)^{-1} - 81 \left(2-0\right)^{-1} \right]\)

\(= \pi \left[ 81 \times 1 - 81 \times \frac{1}{2} \right]\)

\(= \pi \left[ 81 - 40.5 \right]\)

\(= \pi \times 40.5\)

\(= \frac{81\pi}{2}\)

(i) To find the coordinates of \(A\) and \(B\), set \(y = x + \frac{4}{x} = 5\). Solving for \(x\), we get:

\(x + \frac{4}{x} = 5\)

\(x^2 - 5x + 4 = 0\)

\((x-1)(x-4) = 0\)

Thus, \(x = 1\) or \(x = 4\). Therefore, \(A(1, 5)\) and \(B(4, 5)\).

To find \(M\), differentiate \(y = x + \frac{4}{x}\):

\(\frac{dy}{dx} = 1 - \frac{4}{x^2}\)

Set \(\frac{dy}{dx} = 0\) to find the minimum point:

\(1 - \frac{4}{x^2} = 0\)

\(x^2 = 4\)

\(x = 2\)

Substitute \(x = 2\) into \(y = x + \frac{4}{x}\):

\(y = 2 + \frac{4}{2} = 4\)

Thus, \(M(2, 4)\).

(ii) The volume of the solid of revolution is given by:

\(V = \pi \int_{1}^{4} y^2 \, dx\)

\(y = x + \frac{4}{x}\), so \(y^2 = (x + \frac{4}{x})^2 = x^2 + 8 + \frac{16}{x^2}\)

Integrate:

\(\int (x^2 + 8 + \frac{16}{x^2}) \, dx = \frac{x^3}{3} + 8x - \frac{16}{x}\)

Evaluate from 1 to 4:

\(\left[ \frac{x^3}{3} + 8x - \frac{16}{x} \right]_{1}^{4} = \left( \frac{64}{3} + 32 - 4 \right) - \left( \frac{1}{3} + 8 - 16 \right)\)

\(= \left( \frac{64}{3} + 28 \right) - \left( \frac{1}{3} - 8 \right)\)

\(= \frac{64}{3} + 28 + 8 - \frac{1}{3}\)

\(= \frac{63}{3} + 36\)

\(= 21 + 36 = 57\)

Volume of cylinder: \(\pi \times 5^2 \times 3 = 75\pi\)

Volume of solid: \(75\pi - 57\pi = 18\pi\)

The volume of the solid of revolution formed by rotating the curve \(y = \frac{a}{x}\) from \(x = 1\) to \(x = 3\) about the x-axis is given by:

\(V = \pi \int_{1}^{3} \left( \frac{a}{x} \right)^2 \, dx\)

\(= \pi \int_{1}^{3} \frac{a^2}{x^2} \, dx\)

Integrating \(\frac{a^2}{x^2}\) gives:

\(= \pi \left[ -\frac{a^2}{x} \right]_{1}^{3}\)

\(= \pi \left( -\frac{a^2}{3} + \frac{a^2}{1} \right)\)

\(= \pi \left( a^2 - \frac{a^2}{3} \right)\)

\(= \pi \left( \frac{2a^2}{3} \right)\)

We are given that this volume is \(24\pi\), so:

\(\pi \left( \frac{2a^2}{3} \right) = 24\pi\)

\(\frac{2a^2}{3} = 24\)

\(2a^2 = 72\)

\(a^2 = 36\)

\(a = 6\)

To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\)

Given \(f(x) = \frac{3}{2x+5}\), we have:

\([f(x)]^2 = \left(\frac{3}{2x+5}\right)^2 = \frac{9}{(2x+5)^2}\)

The limits of integration are from \(x = 0\) to \(x = 2\).

Thus, the volume \(V\) is:

\(V = \pi \int_{0}^{2} \frac{9}{(2x+5)^2} \, dx\)

Integrating, we have:

\(V = \pi \left[ -\frac{9}{2(2x+5)} \right]_{0}^{2}\)

Evaluating the definite integral:

\(V = \pi \left( -\frac{9}{2(2 \times 2 + 5)} + \frac{9}{2(2 \times 0 + 5)} \right)\)

\(V = \pi \left( -\frac{9}{18} + \frac{9}{10} \right)\)

\(V = \pi \left( -0.5 + 0.9 \right)\)

\(V = \pi \times 0.4\)

\(V = 0.4\pi\)

Thus, the volume is \(0.4\pi\) (or approximately 1.26).

(i) To find the gradient of the curve at \(x = 2\), we first differentiate \(y = \frac{6}{3x - 2}\).

Using the chain rule, \(\frac{dy}{dx} = -6(3x - 2)^{-2} \times 3\).

Substitute \(x = 2\) into the derivative:

\(\frac{dy}{dx} = -6(3 \times 2 - 2)^{-2} \times 3 = -6(4)^{-2} \times 3 = -\frac{9}{8}\).

(ii) To find the volume of the solid of revolution, use the formula:

\(V = \pi \int y^2 \, dx\).

\(V = \pi \int_{1}^{2} \left( \frac{6}{3x - 2} \right)^2 \, dx\).

\(V = \pi \int_{1}^{2} \frac{36}{(3x - 2)^2} \, dx\).

The integral of \(\frac{36}{(3x - 2)^2}\) is \(\frac{-36}{3x - 2}\).

Evaluate from 1 to 2:

\(\left[ \frac{-36}{3x - 2} \right]_{1}^{2} = \left( \frac{-36}{4} - \frac{-36}{1} \right) = -9 + 36 = 27\).

Thus, the volume is \(9\pi\).

(i) To find the area of the shaded region, integrate with respect to \(y\):

\(A = \int_1^2 \left( y^2 - 1 \right) \frac{1}{3} \, dy\)

\(= \left[ \frac{y^3}{9} - \frac{y}{3} \right]_1^2 = \frac{4}{9}\)

(ii) To find the volume when the region is rotated about the \(x\)-axis:

\(V = \pi \int_0^1 y^2 \, dx = \pi \int_0^1 (3x + 1) \, dx\)

\(= \pi \left[ \frac{3x^2}{2} + x \right]_0^1 = \pi \left( \frac{3}{2} + 1 \right) = \frac{5\pi}{2}\)

Subtract the volume of the cylinder: \(\pi \times 2^2 \times 1 = 4\pi\)

\(\text{Volume} = 4\pi - \frac{5\pi}{2} = 1.5\pi\)

(iii) To find the angle between the tangents, calculate the derivatives:

\(\frac{dy}{dx} = \frac{1}{2}(3x + 1)^{-\frac{1}{2}} \times 3\)

At \(x = 0\), \(m = \frac{3}{2}\). At \(x = 1\), \(m = \frac{3}{4}\).

Calculate the angles: \(\arctan(\frac{3}{2}) \approx 56.3^\circ\) and \(\arctan(\frac{3}{4}) \approx 36.9^\circ\).

The angle between the tangents is \(56.3^\circ - 36.9^\circ = 19.4^\circ\).

The volume \(V\) of the solid of revolution about the \(x\)-axis is given by the integral:

\(V = \pi \int_{1}^{4} y^2 \, dx\)

Substitute \(y = 3x^{\frac{1}{4}}\) into the integral:

\(V = \pi \int_{1}^{4} (3x^{\frac{1}{4}})^2 \, dx = \pi \int_{1}^{4} 9x^{\frac{1}{2}} \, dx\)

Integrate \(9x^{\frac{1}{2}}\):

\(\int 9x^{\frac{1}{2}} \, dx = 9 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = 6x^{\frac{3}{2}}\)

Evaluate from 1 to 4:

\(V = \pi \left[ 6x^{\frac{3}{2}} \right]_{1}^{4} = \pi \left( 6 \cdot 4^{\frac{3}{2}} - 6 \cdot 1^{\frac{3}{2}} \right)\)

Calculate \(4^{\frac{3}{2}} = 8\):

\(V = \pi (6 \cdot 8 - 6 \cdot 1) = \pi (48 - 6) = 42\pi\)

The volume \(V\) of the solid of revolution is given by:

\(V = \pi \int_0^1 \left( \frac{6}{5 - 2x} \right)^2 \, dx\)

First, simplify the integrand:

\(\left( \frac{6}{5 - 2x} \right)^2 = \frac{36}{(5 - 2x)^2}\)

Thus, the integral becomes:

\(V = \pi \int_0^1 \frac{36}{(5 - 2x)^2} \, dx\)

Let \(u = 5 - 2x\), then \(du = -2 \, dx\) or \(dx = -\frac{1}{2} \, du\).

Change the limits of integration: when \(x = 0\), \(u = 5\); when \(x = 1\), \(u = 3\).

Substitute and integrate:

\(V = \pi \int_5^3 \frac{36}{u^2} \left(-\frac{1}{2}\right) \, du\)

\(= -18\pi \int_5^3 \frac{1}{u^2} \, du\)

\(= -18\pi \left[ -\frac{1}{u} \right]_5^3\)

\(= 18\pi \left( \frac{1}{3} - \frac{1}{5} \right)\)

\(= 18\pi \left( \frac{5 - 3}{15} \right)\)

\(= 18\pi \left( \frac{2}{15} \right)\)

\(= \frac{36\pi}{15}\)

\(= \frac{12\pi}{5}\)

Thus, the volume obtained is \(\frac{12}{5} \pi\).

The volume of the solid formed by rotating the region about the x-axis is given by the integral:

\(\text{Vol} = \pi \int_{1}^{2} y^2 \, dx\)

Substitute \(y = x^2 + \frac{2}{x}\) into the equation:

\(\text{Vol} = \pi \int_{1}^{2} \left(x^2 + \frac{2}{x}\right)^2 \, dx\)

Expand the integrand:

\(\left(x^2 + \frac{2}{x}\right)^2 = x^4 + 4 + \frac{4x}{x^2}\)

\(= x^4 + 4x + \frac{4}{x^2}\)

Integrate term by term:

\(\int x^4 \, dx = \frac{x^5}{5}\)

\(\int 4x \, dx = 2x^2\)

\(\int \frac{4}{x^2} \, dx = -\frac{4}{x}\)

Thus, the integral becomes:

\(\text{Vol} = \pi \left[ \frac{x^5}{5} - \frac{4}{x} + 2x^2 \right]_{1}^{2}\)

Evaluate the definite integral:

\(\text{Vol} = \pi \left( \left[ \frac{32}{5} - 2 + 8 \right] - \left[ \frac{1}{5} - 4 + 2 \right] \right)\)

\(= \pi \left( \frac{71}{5} \right)\)

Therefore, the volume is \(\frac{71\pi}{5}\) or approximately 44.6.

(i) Differentiate \(y = \frac{18}{x}\) to find the gradient of the tangent: \(\frac{dy}{dx} = -\frac{18}{x^2}\). At \(P(6, 3)\), the gradient is \(-\frac{1}{2}\). The gradient of the normal is the negative reciprocal, \(2\). The equation of the normal is \(y - 3 = 2(x - 6)\), simplifying to \(y = 2x - 9\). Setting \(y = 0\) to find \(R\), we get \(0 = 2x - 9\), so \(x = 4.5\). Thus, \(R\) is \(\left(4\frac{1}{2}, 0\right)\).

(ii) The volume of the solid is given by \(\pi \int_{4.5}^{6} \left(\frac{18}{x}\right)^2 \, dx\). This simplifies to \(\pi \int_{4.5}^{6} \frac{324}{x^2} \, dx = \pi \left[-\frac{324}{x}\right]_{4.5}^{6}\). Evaluating, we get \(\pi \left(-54 + 72\right) = 18\pi\).

(b) To find the volume of revolution, use the formula:

\(V = \pi \int_{1}^{2} y^2 \, dx\)

Substitute \(y = \frac{1}{(3x - 2)^{\frac{3}{2}}}\) into the integral:

\(V = \pi \int_{1}^{2} \left( \frac{1}{(3x - 2)^{\frac{3}{2}}} \right)^2 \, dx = \pi \int_{1}^{2} (3x - 2)^{-3} \, dx\)

Integrate:

\(V = \pi \left[ \frac{(3x - 2)^{-2}}{-2 \times 3} \right]_{1}^{2}\)

\(V = \pi \left[ -\frac{1}{6(3x - 2)^2} \right]_{1}^{2}\)

Evaluate the definite integral:

\(V = \pi \left[ -\frac{1}{6 \times 16} + \frac{1}{6} \right] = \frac{5\pi}{32}\)

(c) Differentiate \(y = \frac{1}{(3x - 2)^{\frac{3}{2}}}\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = -\frac{3}{2} \times 3(3x - 2)^{-\frac{5}{2}}\)

At \(x = 1\), \(\frac{dy}{dx} = -\frac{9}{2}\).

The gradient of the normal is \(-\frac{1}{\left( -\frac{9}{2} \right)} = \frac{2}{9}\).

The equation of the normal is:

\(y - 1 = \frac{2}{9}(x - 1)\)

Substitute \(x = 0\) to find the y-intercept:

\(y - 1 = \frac{2}{9}(0 - 1)\)

\(y = 1 - \frac{2}{9} = \frac{7}{9}\)

(i) To find the equation of the tangent at B, first find the derivative of \(y = \frac{8}{3x + 2}\). The derivative is \(\frac{dy}{dx} = -\frac{24}{(3x + 2)^2}\). At B (2, 1), \(\frac{dy}{dx} = -\frac{3}{8}\).

The equation of the tangent line at B is \(y - 1 = -\frac{3}{8}(x - 2)\).

To find where this line crosses the x-axis, set \(y = 0\):

\(0 - 1 = -\frac{3}{8}(x - 2)\)

\(-1 = -\frac{3}{8}x + \frac{3}{4}\)

\(-\frac{7}{4} = -\frac{3}{8}x\)

\(x = \frac{7}{4} \times \frac{8}{3} = \frac{14}{3}\)

The area of triangle BDC is \(\frac{1}{2} \times 2 \times \frac{2}{3} \times 1 = \frac{4}{3}\).

(ii) The volume of the solid formed by rotating the region ODBA about the x-axis is given by:

\(V = \pi \int_0^2 y^2 \, dx = \pi \int_0^2 \left(\frac{8}{3x + 2}\right)^2 \, dx\)

\(= \pi \int_0^2 \frac{64}{(3x + 2)^2} \, dx\)

Let \(u = 3x + 2\), then \(du = 3 \, dx\), so \(dx = \frac{1}{3} \, du\).

Change the limits: when \(x = 0, u = 2\); when \(x = 2, u = 8\).

\(V = \pi \int_2^8 \frac{64}{u^2} \times \frac{1}{3} \, du\)

\(= \frac{64\pi}{3} \int_2^8 u^{-2} \, du\)

\(= \frac{64\pi}{3} \left[ -u^{-1} \right]_2^8\)

\(= \frac{64\pi}{3} \left( -\frac{1}{8} + \frac{1}{2} \right)\)

\(= \frac{64\pi}{3} \times \frac{3}{8}\)

\(= 8\pi\).

The curve intersects \(y = 1\) at \((3, 1)\).

Volume = \(\pi \int (x-2) \,[dx]\)

\(\left[ \pi \left( \frac{1}{2} x^2 - 2x \right) \right] \text{ or } \left[ \pi \left( \frac{1}{2} (x-2)^2 \right) \right]\)

= \(\left[ \pi \left( \frac{5^2}{2} - 2 \times 5 \right) - \left( \text{their } 3^2/2 - 2 \times \text{their } 3 \right) \right]\)

= \(\left[ \pi \left( \frac{5}{2} + \frac{3}{2} \right) \right] \text{ as a minimum requirement for their values}\)

Volume of cylinder = \(\pi \times 1^2 \times (5 - \text{their } 3) = [2\pi]\)

\([Volume of solid = 4\pi - 2\pi =] 2\pi \text{ or } 6.28\)

(a) To find the volume generated when the shaded region is rotated about the \(y\)-axis, we use the formula for the volume of revolution:

\(\text{Volume} = \pi \int x^2 \, dy\)

Given \(y = \frac{6}{x}\), we have \(x = \frac{6}{y}\). Therefore, \(x^2 = \left(\frac{6}{y}\right)^2 = \frac{36}{y^2}\).

Substitute into the integral:

\(\text{Volume} = \pi \int \frac{36}{y^2} \, dy\)

\(= \pi \left[ -\frac{36}{y} \right]\)

Evaluate from \(y = 2\) to \(y = 6\):

\(= \pi \left( -\frac{36}{6} + \frac{36}{2} \right)\)

\(= \pi ( -6 + 18 )\)

\(= 12\pi\)

Subtract the volume of the cylinder \(\pi \times 1^2 \times 4 = 4\pi\):

\(\text{Total Volume} = 12\pi - 4\pi = 8\pi\)

(b) The gradient of the curve \(y = \frac{6}{x}\) is given by differentiating:

\(\frac{dy}{dx} = -\frac{6}{x^2}\)

The gradient of the line \(y + 2x = 0\) is \(-2\). Set \(-\frac{6}{x^2} = -2\):

\(\frac{6}{x^2} = 2\)

\(x^2 = 3\)

\(x = \sqrt{3}\)

Substitute \(x = \sqrt{3}\) into \(y = \frac{6}{x}\):

\(y = \frac{6}{\sqrt{3}} = 2\sqrt{3}\)

Check if \(y = 2x\):

\(2\sqrt{3} = 2 \times \sqrt{3}\)

Thus, \(X\) lies on \(y = 2x\).

(a) To find the coordinates of \(A\) and \(B\), solve the simultaneous equations:

\(\frac{8}{x+2} = 4 - \frac{1}{2}x\)

Solving gives \(x = 0\) or \(x = 6\).

For \(x = 0\), \(y = \frac{8}{0+2} = 4\), so \(A(0, 4)\).

For \(x = 6\), \(y = \frac{8}{6+2} = 1\), so \(B(6, 1)\).

At \(C\), the derivative of \(y = \frac{8}{x+2}\) is \(-\frac{8}{(x+2)^2}\).

Set \(-\frac{8}{(x+2)^2} = -\frac{1}{2}\) (slope of \(AB\)) and solve for \(x\):

\(-\frac{8}{(x+2)^2} = -\frac{1}{2}\)

\((x+2)^2 = 16\)

\(x = 2\)

Substitute \(x = 2\) into \(y = \frac{8}{x+2}\) to find \(y = 2\), so \(C(2, 2)\).

(b) The volume under the line is:

\(\pi \int_{0}^{6} \left(-\frac{1}{2}x + 4\right)^2 \, dx = \pi \left[ \frac{x^3}{12} - 2x^2 + 16x \right]_{0}^{6} = 42\pi\)

The volume under the curve is:

\(\pi \int_{0}^{6} \left(\frac{8}{x+2}\right)^2 \, dx = \pi \left[ \frac{-64}{x+2} \right]_{0}^{6} = 24\pi\)

The volume of the shaded region is:

\(42\pi - 24\pi = 18\pi\)

To find the volume of the solid obtained by rotating the region about the \(y\)-axis, we use the method of cylindrical shells. The formula for the volume is:

\(V = \pi \int_{1}^{5} (y - 1) \, dy\)

First, integrate \(y - 1\):

\(\int (y - 1) \, dy = \frac{y^2}{2} - y\)

Apply the limits from 1 to 5:

\(V = \pi \left[ \left( \frac{25}{2} - 5 \right) - \left( \frac{1}{2} - 1 \right) \right]\)

Calculate the expression:

\(V = \pi \left[ \frac{25}{2} - 5 - \frac{1}{2} + 1 \right]\)

\(V = \pi \left[ \frac{25}{2} - \frac{1}{2} - 4 \right]\)

\(V = \pi \left[ \frac{24}{2} - 4 \right]\)

\(V = \pi \times 8\)

\(V = 8\pi\)

Thus, the volume obtained is \(8\pi\) or approximately 25.1.