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