Past Exam Questions

The equation of a curve is \(y = \sqrt{(8x - x^2)}\). Find

1. an expression for \(\frac{dy}{dx}\), and the coordinates of the stationary point on the curve,
2. the volume obtained when the region bounded by the curve and the x-axis is rotated through 360° about the x-axis.

The diagram shows the curve \(y = \sqrt{1 + 2x}\) meeting the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\). The \(y\)-coordinate of the point \(C\) on the curve is 3.

1. Find the coordinates of \(B\) and \(C\).
2. Find the equation of the normal to the curve at \(C\).
3. Find the volume obtained when the shaded region is rotated through 360° about the \(y\)-axis.

The diagram shows part of the curve \(y = 4\sqrt{x} - x\). The curve has a maximum point at \(M\) and meets the \(x\)-axis at \(O\) and \(A\).

(i) Find the coordinates of \(A\) and \(M\).

(ii) Find the volume obtained when the shaded region is rotated through 360° about the \(x\)-axis, giving your answer in terms of \(\pi\).

(i) Sketch the curve \(y = (x - 2)^2\).

(ii) The region enclosed by the curve, the \(x\)-axis and the \(y\)-axis is rotated through \(360^\circ\) about the \(x\)-axis. Find the volume obtained, giving your answer in terms of \(\pi\).

The diagram shows part of the curve \(y = \frac{1}{(3x+1)^{\frac{1}{4}}}\). The curve cuts the y-axis at \(A\) and the line \(x = 5\) at \(B\).

(i) Show that the equation of the line \(AB\) is \(y = -\frac{1}{10}x + 1\). [4]

(ii) Find the volume obtained when the shaded region is rotated through 360° about the x-axis. [9]

The diagram shows the circle with equation \((x-2)^2 + y^2 = 8\). The chord \(AB\) of the circle intersects the positive \(y\)-axis at \(A\) and is parallel to the \(x\)-axis.

(a) Find, by calculation, the coordinates of \(A\) and \(B\).

(b) Find the volume of revolution when the shaded segment, bounded by the circle and the chord \(AB\), is rotated through 360° about the \(x\)-axis.

The equation of a curve is \(y = \frac{9}{2-x}\).

Find the volume obtained when the region bounded by the curve, the coordinate axes and the line \(x = 1\) is rotated through 360° about the x-axis.

The diagram shows part of the curve \(y = x + \frac{4}{x}\) which has a minimum point at \(M\). The line \(y = 5\) intersects the curve at the points \(A\) and \(B\).

(i) Find the coordinates of \(A, B\) and \(M\).

(ii) Find the volume obtained when the shaded region is rotated through 360° about the x-axis.

The diagram shows part of the curve \(y = \frac{a}{x}\), where \(a\) is a positive constant. Given that the volume obtained when the shaded region is rotated through 360° about the x-axis is \(24\pi\), find the value of \(a\).

The function \(f\) is such that \(f(x) = \frac{3}{2x+5}\) for \(x \in \mathbb{R}, x \neq -2.5\).

A curve has the equation \(y = f(x)\). Find the volume obtained when the region bounded by the curve, the coordinate axes and the line \(x = 2\) is rotated through 360° about the \(x\)-axis.

The diagram shows part of the curve \(y = \frac{6}{3x - 2}\).

(i) Find the gradient of the curve at the point where \(x = 2\).

(ii) Find the volume obtained when the shaded region is rotated through 360° about the x-axis, giving your answer in terms of \(\pi\).

The diagram shows the curve \(y = \sqrt{3x + 1}\) and the points \(P(0, 1)\) and \(Q(1, 2)\) on the curve. The shaded region is bounded by the curve, the \(y\)-axis and the line \(y = 2\).

(i) Find the area of the shaded region.

(ii) Find the volume obtained when the shaded region is rotated through \(360^\circ\) about the \(x\)-axis.

Tangents are drawn to the curve at the points \(P\) and \(Q\).

(iii) Find the acute angle, in degrees correct to 1 decimal place, between the two tangents.

The diagram shows the curve \(y = 3x^{\frac{1}{4}}\). The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 4\). Find the volume of the solid obtained when this shaded region is rotated completely about the \(x\)-axis, giving your answer in terms of \(\pi\).

The equation of a curve is \(y = \frac{6}{5 - 2x}\).

The region between the curve, the x-axis and the lines \(x = 0\) and \(x = 1\) is rotated through 360° about the x-axis. Show that the volume obtained is \(\frac{12}{5} \pi\).

A curve has equation \(y = x^2 + \frac{2}{x}\).

Find the volume of the solid formed when the region enclosed by the curve, the x-axis and the lines \(x = 1\) and \(x = 2\) is rotated completely about the x-axis.

The diagram shows part of the graph of \(y = \frac{18}{x}\) and the normal to the curve at \(P(6, 3)\). This normal meets the \(x\)-axis at \(R\). The point \(Q\) on the \(x\)-axis and the point \(S\) on the curve are such that \(PQ\) and \(SR\) are parallel to the \(y\)-axis.

(i) Find the equation of the normal at \(P\) and show that \(R\) is the point \(\left(4\frac{1}{2}, 0\right)\).

(ii) Show that the volume of the solid obtained when the shaded region \(PQRS\) is rotated through \(360^\circ\) about the \(x\)-axis is \(18\pi\).

The diagram shows the curve with equation \(y = \frac{1}{(3x - 2)^{\frac{3}{2}}}\). The shaded region is bounded by the curve, the x-axis and the lines \(x = 1\) and \(x = 2\). The shaded region is rotated through 360° about the x-axis.

(b) Find the volume of revolution.

The normal to the curve at the point \((1, 1)\) crosses the y-axis at the point \(A\).

(c) Find the y-coordinate of \(A\).

The diagram shows points A (0, 4) and B (2, 1) on the curve \(y = \frac{8}{3x + 2}\). The tangent to the curve at B crosses the x-axis at C. The point D has coordinates (2, 0).

(i) Find the equation of the tangent to the curve at B and hence show that the area of triangle BDC is \(\frac{4}{3}\).

(ii) Show that the volume of the solid formed when the shaded region ODBA is rotated completely about the x-axis is \(8\pi\).

The diagram shows part of the curve with equation \(y^2 = x - 2\) and the lines \(x = 5\) and \(y = 1\). The shaded region enclosed by the curve and the lines is rotated through 360° about the x-axis.

Find the volume obtained.

The diagram shows part of the curve \(y = \frac{6}{x}\). The points \((1, 6)\) and \((3, 2)\) lie on the curve. The shaded region is bounded by the curve and the lines \(y = 2\) and \(x = 1\).

(a) Find the volume generated when the shaded region is rotated through 360° about the \(y\)-axis. [5]

(b) The tangent to the curve at a point \(X\) is parallel to the line \(y + 2x = 0\). Show that \(X\) lies on the line \(y = 2x\). [3]