Finding the Volume of a Solid

Find the volume \(V\) of the solid under the surface \(z=x^{2}+y^{2}\), above the \(xy\)-plane, and inside the cylinder \(x^{2}+y^{2}=2y\).

Solution See Figure 30. The volume we seek lies above the region \(R\) in the \(xy\)-plane enclosed by the circle \(x^{2}+y^{2}=2y\), whose center is at \((0,1,0)\) and whose radius is 1. To convert to polar coordinates, we let \(x=r\cos \theta\) and \(y=r\sin \theta\). Then the region \(R\) is enclosed by \[ \begin{eqnarray*} x^{2}+y^{2} &=&2y \\[3pt] r^{2}\cos ^{2}\theta +r^{2}\sin ^{2}\theta &=&2r\sin \theta \\[3pt] r^{2} &=&2r\sin \theta \\[3pt] r &=&2\sin \theta \end{eqnarray*} \]

The region \(R\) is bounded by \(0\leq r\leq 2\sin \theta\) and \(0\leq \theta \leq \pi .\)

The volume \(V\) under \(z=x^{2}+y^{2}=r^{2}\), above the \(xy\)-plane, and over \(R\) is \[ \begin{eqnarray*} V &=&\displaystyle\iint\limits_{\kern-3ptR} (x^2+y^2)\,dA =\displaystyle\iint\limits_{\kern-3ptR}r^{2}r\,{\it dr}\,d\theta =\int_{0}^{\pi }\int_{0}^{2\sin \theta }r^{3}\,dr\,d\theta =\int_{0}^{\pi }\left[ \dfrac{r^{4}}{4}\right] _{0}^{2\sin \theta }d\theta \\[4pt] &=&4\int_{0}^{\pi }\sin ^{4}\theta \,d\theta =4\int_{0}^{\pi }\left( \dfrac{ 1-\cos (2\theta) }{2}\right) ^{2} d\theta \\[4pt] &=&\int_{0}^{\pi }[ 1-2\cos (2\theta) +\cos ^{2}(2\theta) ] \,d\theta =\int_{0}^{\pi }\left[ 1-2\cos (2\theta) +\dfrac{1+\cos ( 4\theta ) }{2}\right]\! d\theta \\[4pt] &=&\left[ \theta -\sin (2\theta) +\dfrac{1}{2}\theta +\dfrac{1}{ 2}\cdot \dfrac{\sin ( 4\theta )}{4} \right] _{0}^{\pi }=\dfrac{3\pi }{2} \hbox{ cubic units} \end{eqnarray*} \]