Volumes of revolution about the x-axis Problem set B One question at a time

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

• \(V=\pi\int_{1}^{2}y^{2}\,dx=\pi\int_{1}^{2}\left(3x^{2}\right)^{2}dx\)
• \(\left(3x^{2}\right)^{2}=9 x^{4}\)
• From the calculator: \(\pi\int_{1}^{2}\left(3x^{2}\right)^{2}dx = 175\)

The region between the curve \(y=0.25x^{3}\), 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.

• \(V=\pi\int_{0}^{6}y^{2}\,dx=\pi\int_{0}^{6}\left(0.25x^{3}\right)^{2}dx\)
• \(\left(0.25x^{3}\right)^{2}=\frac{x^{6}}{16}\)
• From the calculator: \(\pi\int_{0}^{6}\left(0.25x^{3}\right)^{2}dx = 7850\)

The region between the curve \(y=x^{2}-5\), the x-axis and the lines \(x=3\) and \(x=6\) is rotated 360° about the x-axis. Find the volume of the solid, to 3 significant figures.

• \(V=\pi\int_{3}^{6}y^{2}\,dx=\pi\int_{3}^{6}\left(x^{2}-5\right)^{2}dx\)
• \(\left(x^{2}-5\right)^{2}=x^{4} - 10 x^{2} + 25\)
• From the calculator: \(\pi\int_{3}^{6}\left(x^{2}-5\right)^{2}dx = 2990\)

The region between the curve \(y=e^{2.5x}\), the x-axis and the lines \(x=0\) and \(x=\ln 2\) is rotated 360° about the x-axis. Find the exact volume of the solid, then give it to 3 significant figures.

• \(V=\pi\int_{0}^{\ln 2}\left(e^{2.5x}\right)^{2}dx=\pi\int_{0}^{\ln 2}e^{5x}\,dx\)
• From the calculator: \(\pi\int_{0}^{\ln 2}e^{5x}\,dx = 19.5\)

The region between the curve \(y=\frac{9}{x+4}\), 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.

• \(V=\pi\int_{0}^{5}\left(\frac{9}{x+4}\right)^{2}dx=\pi\int_{0}^{5}\frac{81}{(x+4)^{2}}\,dx\)
• From the calculator: \(\pi\int_{0}^{5}\frac{81}{(x+4)^{2}}\,dx = 35.3\)

The diagram shows the curve \(y=\sqrt{x}\cos x\) for \(0\le x\le \frac{\pi}{2}\). 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: \(\int_{0}^{\frac{\pi}{2}}\sqrt{x}\cos x\,dx = 0.704\)

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

   • \(V=\pi\int_{0}^{\frac{\pi}{2}}\left(\sqrt{x}\cos x\right)^{2}dx\)
   • From the calculator: \(\pi\int_{0}^{\frac{\pi}{2}}\left(\sqrt{x}\cos x\right)^{2}dx = 1.15\)

A wine glass bowl is a solid of revolution, with lengths in centimetres. The region between the curve \(y=\frac{6}{x+1}\), the x-axis and the lines \(x=1\) and \(x=5\) is rotated 360° about the x-axis. Find the volume of the solid, to 3 significant figures.

• \(V=\pi\int_{1}^{5}\left(\frac{6}{x+1}\right)^{2}dx=\pi\int_{1}^{5}\frac{36}{(x+1)^{2}}\,dx\)
• From the calculator: \(\pi\int_{1}^{5}\frac{36}{(x+1)^{2}}\,dx = 37.7\)

A decorative vase is modelled by rotating a curve about the x-axis, with lengths in decimetres. The region between the curve \(y=e^{x/6}\), the x-axis and the lines \(x=0\) and \(x=\ln 64\) is rotated 360° about the x-axis. Find the exact volume of the solid, then give it to 3 significant figures.

• \(V=\pi\int_{0}^{\ln 64}\left(e^{x/6}\right)^{2}dx=\pi\int_{0}^{\ln 64}e^{x/3}\,dx\)
• From the calculator: \(\pi\int_{0}^{\ln 64}e^{x/3}\,dx = 28.3\)

The side view of a wooden finial is modelled with \(x\) and \(y\) in cm. From \(x=0\) to \(x=3\) the edge is the line \(y=2.5+\frac{x}{2}\). From \(x=3\) to \(x=9\) the edge is a quadratic with its maximum point at \((3,\,4)\), which passes through \((9,\,0)\). The finial is the solid made by rotating the region under the edge 360° about the x-axis. Find its volume, to 3 significant figures.

• Quadratic: \(y=4+a(x-3)^2\) and \(y=0\) at \(x=9\), so \(4+36a=0\) and \(a=-\frac{1}{9}\)
• \(V=\pi\int_0^{3}\left(2.5+\frac{x}{2}\right)^2dx+\pi\int_{3}^{9}\left(4-\frac{1}{9}(x-3)^2\right)^2dx\)
• From the calculator: first piece 101.3, second piece 160.8
• V = 101.3 + 160.8 = 262.2 = 262 cm³ (3 s.f.)

A candle mould is made by rotating a curve about the x-axis, with lengths in centimetres. The region between the curve \(y=a x^{3}\), the x-axis and the line \(x=2\), where \(a>0\), is rotated 360° about the x-axis. The volume of the solid is 150 cm³. Find the value of a.

• \(V=\pi\int_{0}^{2}\left(ax^{3}\right)^{2}dx=\pi a^{2}\int_{0}^{2}x^{6}\,dx=\pi a^{2}\times\frac{128}{7}\)
• \(\frac{128}{7}\pi a^{2}=150\Rightarrow a^{2}=2.61\)
• a = 1.62

A lamp base is made by rotating a curve about the x-axis, with lengths in centimetres. The region between the curve \(y=\frac{8}{x+1}\), the x-axis and the lines \(x=1\) and \(x=3\) is rotated 360° about the x-axis. Show that the volume of the solid is exactly \(16\pi\).

• \(V=\pi\int_{1}^{3}\left(\frac{8}{x+1}\right)^{2}dx=\pi\int_{1}^{3}\frac{64}{(x+1)^{2}}\,dx\)
• \(\int \frac{64}{(x+1)^{2}}dx=-\frac{64}{x+1}\)
• \(V=\pi\left[-\frac{64}{x+1}\right]_{1}^{3}=\pi\left(32-16\right)=16\pi\)