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

The part of the curve \(y=2x^{2}+3\) for \(0\le x\le 3\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=2x^{2}+3\Rightarrow x^{2}=\frac{y-3}{2}\)
• Limits: \(x=0\) gives \(y=3\), and \(x=3\) gives \(y=21\)
• \(V=\pi\int_{3}^{21}\frac{y-3}{2}\,dy\)
• From the calculator: \(\pi\int_{3}^{21}\frac{y-3}{2}\,dy = 254\)

The part of the curve \(y=2+\ln x\) for \(1\le x\le 4\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=2+\ln x\Rightarrow \ln x=y-2\Rightarrow x=e^{y-2}\), so \(x^{2}=e^{2y-4}\)
• Limits: \(x=1\) gives \(y=2\), and \(x=4\) gives \(y=2+\ln 4\)
• \(V=\pi\int_{2}^{2+\ln 4}e^{2y-4}\,dy\)
• From the calculator: \(\pi\int_{2}^{2+\ln 4}e^{2y-4}\,dy = 23.6\)

The part of the curve \(y=\tan x\) for \(0\le x\le 1\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve. Use radians.

• \(y=\tan x\Rightarrow x=\tan^{-1}y\), so \(x^{2}=\left(\tan^{-1}y\right)^{2}\)
• Limits: \(x=0\) gives \(y=0\), and \(x=1\) gives \(y=\tan 1=1.56\)
• Radian mode.
• From the calculator: \(\pi\int_{0}^{\tan 1}\left(\tan^{-1}y\right)^{2}dy = 2.20\)

The part of the curve \(y=\frac{2}{x-3}\) for \(4\le x\le 8\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=\frac{2}{x-3}\Rightarrow x-3=\frac{2}{y}\Rightarrow x=\frac{2}{y}+3\)
• \(x^{2}=\left(\frac{2}{y}+3\right)^{2}=\frac{4}{y^{2}}+\frac{12}{y}+9\)
• Limits: \(x=4\) gives \(y=2\), and \(x=8\) gives \(y=\frac{2}{5}\)
• From the calculator: \(\pi\int_{0.4}^{2}\left(\frac{2}{y}+3\right)^{2}dy = 131\)

The diagram shows the region bounded by the curve \(y=\frac{2}{x}\), the y-axis and the lines \(y=1\) and \(y=4\).

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

   • \(x=\frac{2}{y}\)
   • Area = \(\int_{1}^{4}\frac{2}{y}\,dy\)
   • From the calculator: \(\int_{1}^{4}\frac{2}{y}\,dy = 2.77\)

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

   • \(x^{2}=\frac{4}{y^{2}}\)
   • \(V=\pi\int_{1}^{4}\frac{4}{y^{2}}\,dy=\pi\left[-\frac{4}{y}\right]_{1}^{4}\)
   • \(V=3\pi\)

The region between the curve \(y=\sqrt{x^{2}+4}\), the y-axis and the line \(y=5\) is shown in the diagram. It is rotated 360° about the y-axis. Find the volume of the solid, as a multiple of π.

• \(y^{2}=x^{2}+4\Rightarrow x^{2}=y^{2}-4\)
• At \(x=0\), \(y=\sqrt{4}=2\). The region runs from \(y=2\) to \(y=5\).
• \(V=\pi\int_{2}^{5}\left(y^{2}-4\right)dy\)
• \(V=\pi\left[\frac{y^{3}}{3}-4y\right]_{2}^{5}=27\pi\)

A bowl is made by rotating a curve about the y-axis, with lengths in centimetres. The part of the curve \(y=\frac{x^{2}}{2}+2\) for \(0\le x\le 4\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=\frac{x^{2}}{2}+2\Rightarrow x^{2}=2\left(y-2\right)\)
• Limits: \(x=0\) gives \(y=2\), and \(x=4\) gives \(y=10\)
• \(V=\pi\int_{2}^{10}2\left(y-2\right)\,dy\)
• From the calculator: \(\pi\int_{2}^{10}2\left(y-2\right)\,dy = 201\)

A vase is made by rotating a curve about the y-axis, with lengths in centimetres. The part of the curve \(y=3+\ln x\) for \(1\le x\le 5\) is rotated 360° about the y-axis. Find the volume of the solid enclosed by the curve, the y-axis and the horizontal lines through the two ends of the curve.

• \(y=3+\ln x\Rightarrow \ln x=y-3\Rightarrow x=e^{y-3}\), so \(x^{2}=e^{2y-6}\)
• Limits: \(x=1\) gives \(y=3\), and \(x=5\) gives \(y=3+\ln 5\)
• \(V=\pi\int_{3}^{3+\ln 5}e^{2y-6}\,dy\)
• From the calculator: \(\pi\int_{3}^{3+\ln 5}e^{2y-6}\,dy = 37.7\)

A water jug is made by rotating the curve \(y=9\ln(x+1)\), for \(0\le y\le 18\), 360° about the y-axis. Lengths are in cm.

1. (a) Find the capacity of the jug in litres.

   • \(y=9\ln(x+1)\Rightarrow x=e^{y/9}-1\)
   • \(V=\pi\int_0^{18}\left(e^{y/9}-1\right)^2dy\)
   • From the calculator: V = 453 cm³ = 0.453 litres

2. (b) The jug holds 300 cm³ of water. Find the depth of the water.

   • Solve \(\pi\int_0^{h}\left(e^{y/9}-1\right)^2dy=300\) for h
   • From the calculator: h = 16.6 cm
   • This is less than 18, so it fits

The diagram shows the region bounded by the curve \(x=4-(y-2)^{2}\) and the line \(x=3\). The region is rotated 360° about the y-axis. Find the volume of the solid, as a multiple of π.

• Curve meets line: \(4-(y-2)^{2}=3\Rightarrow (y-2)^{2}=1\), so \(y=1\) and \(y=3\)
• Outer radius \(x=4-(y-2)^{2}\) (the curve), inner radius \(x=3\) (the line)
• \(V=\pi\int_{1}^{3}\left[\left(4-(y-2)^{2}\right)^{2}-3^{2}\right]dy\)
• From the calculator: \(\pi\int_{1}^{3}\left[\left(4-(y-2)^{2}\right)^{2}-3^{2}\right]dy = 28.5\)
• Exact value: \(\frac{136\pi}{15}\)

A paperweight is made by rotating a curve about the y-axis. The part of the curve \(y=x^{2}+1\) for \(0\le x\le k\) is rotated 360° about the y-axis. The solid enclosed by the curve, the y-axis and the horizontal lines through the ends of the curve has volume 50 cm³. Find the value of k.

• \(x^{2}=y-1\). When \(x=0\), \(y=1\). When \(x=k\), \(y=k^{2}+1\).
• \(V=\pi\int_{1}^{k^{2}+1}(y-1)\,dy=\pi\left[\frac{(y-1)^{2}}{2}\right]_{1}^{k^{2}+1}=\frac{\pi k^{4}}{2}\)
• \(\frac{\pi k^{4}}{2}=50\Rightarrow k^{4}=31.8\)
• k = 2.38