4.08d Volumes of revolution: about x and y axes

\includegraphics{figure_7} The diagram shows the curve with equation \(y = 2x - e^{\frac{1}{2}x}\). The shaded region is bounded by the curve, the \(x\)-axis and the lines \(x = 2\) and \(x = 4\).

1. Find the area of the shaded region, giving your answer in terms of e. [4]

The shaded region is rotated through four right angles about the \(x\)-axis.

1. Using Simpson's rule with two strips, estimate the volume of the solid formed. [5]

1. Use integration by parts to show that $$\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx = \frac{1}{4}\pi - \frac{1}{2} \ln 2.$$ [6]

\includegraphics{figure_1} The finite region \(R\), bounded by the equation \(y = x^{\frac{1}{2}} \sec x\), the line \(x = \frac{\pi}{4}\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.

1. Find the volume of the solid of revolution generated. [2]
2. Find the gradient of the curve with equation \(y = x^{\frac{1}{2}} \sec x\) at the point where \(x = \frac{\pi}{4}\). [3]

Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2t^2, y = 4t, \quad -\sqrt{2} < t < \sqrt{2}.$$ P\((2t^2, 4t)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P, and PR is the line through P parallel to the \(x\)-axis. Q is the point (2, 0). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \includegraphics{figure_8}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac{1}{t}\). [3]
2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2\theta\), and that angle TPQ is equal to \(\theta\). [8]

[The above result shows that if a lamp bulb is placed at Q, then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.

1. Show that the curve has cartesian equation \(y^2 = 8x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\). [7]

The part of the curve \(y = 4 - x^2\) that is above the \(x\)-axis is rotated about the \(y\)-axis. This is shown in Fig. 4. Find the volume of revolution produced, giving your answer in terms of \(\pi\). [5] \includegraphics{figure_4}

Fig. 6 shows the region enclosed by part of the curve \(y = 2x^2\), the straight line \(x + y = 3\), and the \(y\)-axis. The curve and the straight line meet at P (1, 2). \includegraphics{figure_6} The shaded region is rotated through \(360°\) about the \(y\)-axis. Find, in terms of \(\pi\), the volume of the solid of revolution formed. [7] [You may use the formula \(V = \frac{1}{3}\pi r^2 h\) for the volume of a cone.]

Fig. 7a shows the curve with the parametric equations $$x = 2\cos\theta, \quad y = \sin 2\theta, \quad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}.$$ The curve meets the \(x\)-axis at O and P. Q and R are turning points on the curve. The scales on the axes are the same. \includegraphics{figure_7a}

1. State, with their coordinates, the points on the curve for which \(\theta = -\frac{\pi}{2}\), \(\theta = 0\) and \(\theta = \frac{\pi}{2}\). [3]
2. Find \(\frac{dy}{dx}\) in terms of \(\theta\). Hence find the gradient of the curve when \(\theta = \frac{\pi}{2}\), and verify that the two tangents to the curve at the origin meet at right angles. [5]
3. Find the exact coordinates of the turning point Q. [3]

When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \includegraphics{figure_7b}

1. Express \(\sin^2\theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y^2 = x^2(1 - \frac{1}{4}x^2)\). [4]
2. Find the volume of the paperweight shape. [4]

Fig. 7 shows the curve BC defined by the parametric equations $$x = 5 \ln u, \quad y = u + \frac{1}{u}, \quad 1 \leq u \leq 10.$$ The point A lies on the \(x\)-axis and AC is parallel to the \(y\)-axis. The tangent to the curve at C makes an angle \(\theta\) with AC, as shown. \includegraphics{figure_7}

1. Find the lengths OA, OB and AC. [5]
2. Find \(\frac{dy}{dx}\) in terms of \(u\). Hence find the angle \(\theta\). [6]
3. Show that the cartesian equation of the curve is \(y = e^{x/5} + e^{-x/5}\). [2]

An object is formed by rotating the region OACB through \(360°\) about Ox.

1. Find the volume of the object. [5]

Fig. 6 shows the region enclosed by the curve \(y = (1 + 2x^2)^{\frac{1}{2}}\) and the line \(y = 2\). \includegraphics{figure_6} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\). [6]

\includegraphics{figure_6} Figure 1 shows the curve with equation \(y = x\sqrt{1-x}\), \(0 \leq x \leq 1\).

1. Use the substitution \(u^2 = 1 - x\) to show that the area of the region bounded by the curve and the \(x\)-axis is \(\frac{8}{15}\). [8]
2. Find, in terms of \(\pi\), the volume of the solid formed when the region bounded by the curve and the \(x\)-axis is rotated through \(360°\) about the \(x\)-axis. [5]

1. Find \(\int \tan^2 x \, dx\). [3]
2. Show that $$\int \tan x \, dx = \ln|\sec x| + c,$$ where \(c\) is an arbitrary constant. [4]

\includegraphics{figure_1} Figure 1 shows part of the curve with equation \(y = x^2 \tan x\). The shaded region bounded by the curve, the \(x\)-axis and the line \(x = \frac{\pi}{3}\) is rotated through \(2\pi\) radians about the \(x\)-axis.

1. Show that the volume of the solid formed is \(\frac{1}{18}\pi^2(6\sqrt{3} - \pi) - \pi \ln 2\). [6]

\includegraphics{figure_1} Figure 1 shows the curve with equation \(y = 2\sin x + \cosec x\), \(0 < x < \pi\). The shaded region bounded by the curve, the \(x\)-axis and the lines \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\) is rotated through \(360°\) about the \(x\)-axis. Show that the volume of the solid formed is \(\frac{1}{2}\pi(4\pi + 3\sqrt{3})\). [8]

The region bounded by the curve \(y = x^2 - 2x\) and the \(x\)-axis is rotated through \(2\pi\) radians about the \(x\)-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi\). [6]

\includegraphics{figure_1} Figure 1 shows the curve with equation \(y = x\sqrt{\ln x}\), \(x \geq 1\). The shaded region is bounded by the curve, the \(x\)-axis and the line \(x = 3\).

1. Using the trapezium rule with two intervals of equal width, estimate the area of the shaded region. [4]

The shaded region is rotated through \(360°\) about the \(x\)-axis.

1. Find the exact volume of the solid formed. [7]

1. Find \(\int \tan^2 x \, dx\). [3]
2. Show that $$\int \tan x \, dx = \ln |\sec x| + c,$$ where \(c\) is an arbitrary constant. [4]

\includegraphics{figure_8} The diagram shows part of the curve with equation \(y = x^{\frac{1}{2}} \tan x\). The shaded region bounded by the curve, the \(x\)-axis and the line \(x = \frac{\pi}{3}\) is rotated through \(360°\) about the \(x\)-axis.

1. Show that the volume of the solid formed is \(\frac{1}{18}\pi^2(6\sqrt{3} - \pi) - \pi \ln 2\). [5]

A curve is defined parametrically by $$x = t^3 + 5, \quad y = 6t^2 - 1$$ The arc length between the points where \(t = 0\) and \(t = 3\) on the curve is \(s\).

1. Show that \(s = \int_{0}^{3} 3t\sqrt{t^2 + A} \, \text{d}t\), stating the value of the constant \(A\). [4 marks]
2. Hence show that \(s = 61\). [4 marks]

A hyperbola \(H\) has the equation $$\frac{x^2}{a^2} - \frac{y^2}{4a^2} = 1$$ where \(a\) is a positive constant.

1. Write down the equations of the asymptotes of \(H\). [1 mark]
2. Sketch the hyperbola \(H\) on the axes below, indicating the coordinates of any points of intersection with the coordinate axes. The asymptotes have already been drawn. [2 marks]
3. The finite region bounded by \(H\), the positive \(x\)-axis, the positive \(y\)-axis and the line \(y = a\) is rotated through \(360°\) about the \(y\)-axis. Show that the volume of the solid generated is \(ma^3\), where \(m = 3.40\) correct to three significant figures. [5 marks]

A segment of the line \(y = kx\) is rotated about the \(x\)-axis to generate a cone with vertex \(O\). The distance of \(O\) from the centre of the base of the cone is \(h\). The radius of the base of the cone is \(r\). \includegraphics{figure_15}

1. Find \(k\) in terms of \(r\) and \(h\). [1 mark]
2. Use calculus to prove that the volume of the cone is $$\frac{1}{3}\pi r^2 h$$ [3 marks]

The hyperbola \(H\) has equation \(y^2 - x^2 = 16\) The circle \(C\) has equation \(x^2 + y^2 = 32\) The diagram below shows part of the graph of \(H\) and part of the graph of \(C\). \includegraphics{figure_14} Show that the shaded region in the first quadrant enclosed by \(H\), \(C\), the \(x\)-axis and the \(y\)-axis has area $$\frac{16\pi}{3} + 8\ln\left(\frac{\sqrt{2} + \sqrt{6}}{2}\right)$$ [12 marks]

A parabola \(P_1\) has equation \(y^2 = 4ax\) where \(a > 0\) \(P_1\) is translated by the vector \(\begin{bmatrix} b \\ 0 \end{bmatrix}\), where \(b > 0\), to give the parabola \(P_2\)

1. The line \(y = mx\) is a tangent to \(P_2\) Prove that \(m = \pm\sqrt{\frac{a}{b}}\) Solutions using differentiation will be given no marks. [4 marks]
2. The line \(y = \sqrt{\frac{a}{b}} x\) meets \(P_2\) at the point \(D\). The finite region \(R\) is bounded by the \(x\)-axis, \(P_2\) and a line through \(D\) perpendicular to the \(x\)-axis. The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis to form a solid. Find, in terms of \(a\) and \(b\), the volume of this solid. Fully justify your answer. [5 marks]

The diagram shows part of the graph of \(y = \cos^{-1} x\) \includegraphics{figure_7} The finite region enclosed by the graph of \(y = \cos^{-1} x\), the \(y\)-axis, the \(x\)-axis and the line \(x = 0.8\) is rotated by \(2\pi\) radians about the \(x\)-axis. Use Simpson's rule with five ordinates to estimate the volume of the solid formed. Give your answer to four decimal places. [5 marks]

The function f is defined by $$f(x) = \frac{ax + 5}{x + b}$$ where \(a\) and \(b\) are constants. The graph of \(y = f(x)\) has asymptotes \(x = -2\) and \(y = 3\)

1. Write down the value of \(a\) and the value of \(b\) [2 marks]
2. The diagram shows the graph of \(y = f(x)\) and its asymptotes. The shaded region \(R\) is enclosed by the graph of \(y = f(x)\), the \(x\)-axis and the \(y\)-axis. \includegraphics{figure_16}
   1. The shaded region \(R\) is rotated through \(360°\) about the \(x\)-axis to form a solid. Find the volume of this solid. Give your answer to three significant figures. [3 marks]
   2. The shaded region \(R\) is rotated through \(360°\) about the \(y\)-axis to form a solid. Find the volume of this solid. Give your answer to three significant figures. [4 marks]

\(O\) is the origin of a coordinate system whose units are cm. The points \(A\), \(B\), \(C\) and \(D\) have coordinates \((1, 0)\), \((1, 4)\), \((6, 9)\) and \((0, 9)\) respectively. The arc \(BC\) is part of the curve with equation \(x^2 + (y - 10)^2 = 37\). The closed shape \(OABCD\) is formed, in turn, from the line segments \(OA\) and \(AB\), the arc \(BC\) and the line segments \(CD\) and \(DO\) (see diagram). A funnel can be modelled by rotating \(OABCD\) by \(2\pi\) radians about the \(y\)-axis. \includegraphics{figure_6} Find the volume of the funnel according to the model. [3]