Volumes of revolution about the x-axis Problem set A Worked solutions

The region between the curve \(y=2x^{2}\), the x-axis and the lines \(x=1\) and \(x=3\) is rotated 360° about the x-axis. Find the volume of the solid, to 3 significant figures.

• \(V=\pi\int_{1}^{3}y^{2}\,dx=\pi\int_{1}^{3}\left(2x^{2}\right)^{2}dx\)
• \(\left(2x^{2}\right)^{2}=4 x^{4}\)
• From the calculator: \(\pi\int_{1}^{3}\left(2x^{2}\right)^{2}dx = 608\)

The region between the curve \(y=0.5x^{3}\), the x-axis and the lines \(x=0\) and \(x=4\) is rotated 360° about the x-axis. Find the volume of the solid, to 3 significant figures.

• \(V=\pi\int_{0}^{4}y^{2}\,dx=\pi\int_{0}^{4}\left(0.5x^{3}\right)^{2}dx\)
• \(\left(0.5x^{3}\right)^{2}=\frac{x^{6}}{4}\)
• From the calculator: \(\pi\int_{0}^{4}\left(0.5x^{3}\right)^{2}dx = 1840\)

The region between the curve \(y=x^{2}-3\), the x-axis and the lines \(x=2\) and \(x=5\) is rotated 360° about the x-axis. Find the volume of the solid, to 3 significant figures.

• \(V=\pi\int_{2}^{5}y^{2}\,dx=\pi\int_{2}^{5}\left(x^{2}-3\right)^{2}dx\)
• \(\left(x^{2}-3\right)^{2}=x^{4} - 6 x^{2} + 9\)
• From the calculator: \(\pi\int_{2}^{5}\left(x^{2}-3\right)^{2}dx = 1290\)

The region between the curve \(y=e^{1.5x}\), the x-axis and the lines \(x=0\) and \(x=\ln 4\) is rotated 360° about the x-axis. Find the exact volume of the solid, then give it to 3 significant figures.

• \(V=\pi\int_{0}^{\ln 4}\left(e^{1.5x}\right)^{2}dx=\pi\int_{0}^{\ln 4}e^{3x}\,dx\)
• From the calculator: \(\pi\int_{0}^{\ln 4}e^{3x}\,dx = 66.0\)

The region between the curve \(y=\frac{4}{x+3}\), the x-axis and the lines \(x=0\) and \(x=5\) is rotated 360° about the x-axis. Find the volume of the solid, to 3 significant figures.

• \(V=\pi\int_{0}^{5}\left(\frac{4}{x+3}\right)^{2}dx=\pi\int_{0}^{5}\frac{16}{(x+3)^{2}}\,dx\)
• From the calculator: \(\pi\int_{0}^{5}\frac{16}{(x+3)^{2}}\,dx = 10.5\)

The diagram shows the curve \(y=\sqrt{x}\sin\frac{x}{2}\) for \(0\le x\le \pi\). The region between the curve and the x-axis is shaded.

1. (a) Find the area of the shaded region.

   • Radian mode.
   • From the calculator: \(\int_{0}^{\pi}\sqrt{x}\sin\frac{x}{2}\,dx = 2.76\)

2. (b) The region is rotated 360° about the x-axis. Find the volume of the solid formed.

   • \(V=\pi\int_{0}^{\pi}\left(\sqrt{x}\sin\frac{x}{2}\right)^{2}dx\)
   • From the calculator: \(\pi\int_{0}^{\pi}\left(\sqrt{x}\sin\frac{x}{2}\right)^{2}dx = 10.9\)

A trumpet bell is a solid of revolution, with lengths in centimetres. The region between the curve \(y=\frac{8}{x+2}\), the x-axis and the lines \(x=0\) and \(x=6\) is rotated 360° about the x-axis. Find the volume of the solid, to 3 significant figures.

• \(V=\pi\int_{0}^{6}\left(\frac{8}{x+2}\right)^{2}dx=\pi\int_{0}^{6}\frac{64}{(x+2)^{2}}\,dx\)
• From the calculator: \(\pi\int_{0}^{6}\frac{64}{(x+2)^{2}}\,dx = 75.4\)

A horn is modelled by rotating a curve about the x-axis, with lengths in metres. The region between the curve \(y=e^{x/4}\), the x-axis and the lines \(x=0\) and \(x=\ln 25\) is rotated 360° about the x-axis. Find the exact volume of the solid, then give it to 3 significant figures.

• \(V=\pi\int_{0}^{\ln 25}\left(e^{x/4}\right)^{2}dx=\pi\int_{0}^{\ln 25}e^{x/2}\,dx\)
• From the calculator: \(\pi\int_{0}^{\ln 25}e^{x/2}\,dx = 25.1\)

The side view of a toy rocket is modelled with \(x\) and \(y\) in cm. From \(x=0\) to \(x=4\) the edge is the line \(y=2+\frac{3x}{4}\). From \(x=4\) to \(x=8\) the edge is a quadratic with its maximum point at \((4,\,5)\), which passes through \((8,\,0)\). The rocket is the solid made by rotating the region under the edge 360° about the x-axis. Find its volume, to 3 significant figures.

• Quadratic: \(y=5+a(x-4)^2\) and \(y=0\) at \(x=8\), so \(5+16a=0\) and \(a=-\frac{5}{16}\)
• \(V=\pi\int_0^{4}\left(2+\frac{3x}{4}\right)^2dx+\pi\int_{4}^{8}\left(5-\frac{5}{16}(x-4)^2\right)^2dx\)
• From the calculator: first piece 163.4, second piece 167.6
• V = 163.4 + 167.6 = 330.9 = 331 cm³ (3 s.f.)

A bowl is made by rotating a curve about the x-axis, with lengths in centimetres. The region between the curve \(y=a x^{2}\), the x-axis and the line \(x=2\), where \(a>0\), is rotated 360° about the x-axis. The volume of the solid is 90 cm³. Find the value of a.

• \(V=\pi\int_{0}^{2}\left(ax^{2}\right)^{2}dx=\pi a^{2}\int_{0}^{2}x^{4}\,dx=\pi a^{2}\times\frac{32}{5}\)
• \(\frac{32}{5}\pi a^{2}=90\Rightarrow a^{2}=4.48\)
• a = 2.12

A glass is made by rotating a curve about the x-axis, with lengths in centimetres. The region between the curve \(y=\frac{6}{x+3}\), the x-axis and the lines \(x=0\) and \(x=3\) is rotated 360° about the x-axis. Show that the volume of the solid is exactly \(6\pi\).

• \(V=\pi\int_{0}^{3}\left(\frac{6}{x+3}\right)^{2}dx=\pi\int_{0}^{3}\frac{36}{(x+3)^{2}}\,dx\)
• \(\int \frac{36}{(x+3)^{2}}dx=-\frac{36}{x+3}\)
• \(V=\pi\left[-\frac{36}{x+3}\right]_{0}^{3}=\pi\left(12-6\right)=6\pi\)