Volume of revolution with substitution

A question is this type if and only if it involves rotating a region about an axis to find volume, where the integration requires substitution.

6 questions · Challenging +1.0

1.08h Integration by substitution4.08d Volumes of revolution: about x and y axes
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CAIE P1 2021 November Q10
12 marks Standard +0.3
10
  1. Find \(\int _ { 1 } ^ { \infty } \frac { 1 } { ( 3 x - 2 ) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x\). \includegraphics[max width=\textwidth, alt={}, center]{af7aeda9-2ded-4db4-9ff3-ed6adc67859f-16_499_689_1322_726} The diagram shows the curve with equation \(y = \frac { 1 } { ( 3 x - 2 ) ^ { \frac { 3 } { 2 } } }\). 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 \(360 ^ { \circ }\) about the \(x\)-axis.
  2. Find the volume of revolution.
    The normal to the curve at the point \(( 1,1 )\) crosses the \(y\)-axis at the point \(A\).
  3. Find the \(y\)-coordinate of \(A\).
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Edexcel P4 2024 January Q7
8 marks Challenging +1.2
  1. (a) Using the substitution \(u = 4 x + 2 \sin 2 x\), or otherwise, show that
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { 4 x + 2 \sin 2 x } \cos ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 8 } \left( \mathrm { e } ^ { 2 \pi } - 1 \right)$$ Figure 3 The curve shown in Figure 3, has equation $$y = 6 \mathrm { e } ^ { 2 x + \sin 2 x } \cos x$$ The region \(R\), shown shaded in Figure 3, is bounded by the positive \(x\)-axis, the positive \(y\)-axis and the curve. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid.
(b) Use the answer to part (a) to find the volume of the solid formed, giving the answer in simplest form.
Pre-U Pre-U 9794/2 2018 June Q10
10 marks Challenging +1.2
10
  1. By using the substitution \(u = 3 - 2 x\), or otherwise, show that \(\int _ { 0 } ^ { 1 } \left( \frac { 4 x } { 3 - 2 x } \right) ^ { 2 } \mathrm {~d} x = 16 - 12 \ln 3\).
  2. \includegraphics[max width=\textwidth, alt={}, center]{f4b66aaa-16b9-4b15-b3f5-b9657fe98274-4_595_588_927_817} The diagram shows the region \(R\), which is bounded by the curve \(y = \frac { 4 x } { 3 - 2 x }\), the \(y\)-axis and the line \(y = 4\). Find the exact volume generated when the region \(R\) is rotated completely around the \(x\)-axis. {www.cie.org.uk} after the live examination series. }
Edexcel C4 Q6
13 marks Standard +0.8
\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]
Pre-U Pre-U 9794/2 2012 June Q10
12 marks Standard +0.3
    1. Find \(\int \frac{e^x}{1 + e^x} dx\). [2]
    2. Hence evaluate \(\int_0^{\ln 3} \frac{e^x}{1 + e^x} dx\), giving your answer in the form \(\ln k\), where \(k\) is an integer. [3]
    1. Using the substitution \(u = 1 + e^x\), find \(\int \left(\frac{e^x}{1 + e^x}\right)^2 dx\). [5]
    2. Hence find the exact volume of the solid of revolution generated when the curve given by \(y = \frac{e^x}{1 + e^x}\), between \(x = -\ln 3\) and \(x = \ln 3\), is rotated through \(2\pi\) radians about the \(x\)-axis. [2]
Edexcel AEA 2015 June Q7
19 marks Hard +2.3
  1. Use the substitution \(x = \sec\theta\) to show that $$\int_{\sqrt{2}}^{2} \frac{1}{(x^2 - 1)^{\frac{3}{2}}} \, dx = \frac{\sqrt{6} - 2}{\sqrt{3}}$$ [5]
  2. Use integration by parts to show that $$\int \cos\theta \cot^2\theta \, d\theta = \frac{1}{2}[\ln|\cos\theta + \cot\theta| - \cos\theta \cot\theta] + c$$ [6] % Figure shows a curve y = 1/(x^2-1)^(1/2) for x > 1, with shaded region R between x = sqrt(2) and x = 2 \includegraphics{figure_2} Figure 2 shows a sketch of part of the curve with equation \(y = \frac{1}{(x^2 - 1)^{\frac{1}{2}}}\) for \(x > 1\) The region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the lines \(x = \sqrt{2}\) and \(x = 2\) The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.
  3. Show that the volume of the solid formed is $$\pi \left[\frac{3}{8}\ln\left(\frac{1 + \sqrt{2}}{\sqrt{3}}\right) + \frac{7}{36} - \frac{\sqrt{2}}{8}\right]$$ [8]