Finding a Triple Integral Using Cylindrical Coordinates
Give a geometric interpretation of the triple integral \begin{equation*} \int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{0}^{2\sqrt{ 1-x^{2}-y^{2}}}\,{\it dz}\,{\it dy}\,{\it dx} \end{equation*}
Then use cylindrical coordinates to find the triple integral.
The limits of integration on \(z\) are \(z=0\) and \(z=2\sqrt{1-x^{2}-y^{2}}= \sqrt{4-4x^{2}-4y^{2}}\) or, equivalently, \(z^{2}=4-4x^{2}-4y^{2},\) \(z\geq 0.\) We can interpret the integral as the volume of a solid \(E\) that is the upper half of the ellipsoid \(4x^{2}+4y^{2}+z^{2}=4,\) \(z\geq 0,\) as shown in Figure 55. From the \(x\) and \(y\) limits of integration, the projection onto the \(xy\)-plane is the region \(R\) enclosed by the circle \(x^{2}+y^{2}=1\).
To find the integral, we convert the rectangular coordinates to cylindrical coordinates. Then \[ z=2\sqrt{1-x^{2}-y^{2}}=2\sqrt{1-(x^{2}+y^{2}) }=2\sqrt{1-r^{2}} \]
The projection onto the \(xy\)-plane is the region enclosed by the circle \(x^{2}+y^{2}=1.\) So in cylindrical coordinates, we have \[ 0\leq z\leq 2\sqrt{1-r^{2}} \qquad 0\leq r\leq 1\qquad 0\leq \theta \leq 2\pi \]
Then \[ \begin{array}{lll} \int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{0}^{2\sqrt{ 1-x^{2}-y^{2}}}\,{\it dz}\,{\it dy}\,{\it dx}& =&\int_{0}^{2\pi }\int_{0}^{1}\int_{0}^{2\sqrt{ 1-r^{2}}}\,{\it dz}\,r\,dr\,d\theta\\ &=&\int_{0}^{2\pi }\int_{0}^{1} \big[ z\big] _{0}^{2\sqrt{1-r^{2}}}r\,dr\,d\theta =\int_{0}^{2\pi }\int_{0}^{1}2r\sqrt{1-r^{2}}\ dr\,d\theta \\[5pt] & =&\int_{0}^{2\pi }\left[ -\dfrac{2}{3}(1-r^{2})^{3/2}\right] _{0}^{1}\,d\theta =\dfrac{2}{3}\int_{0}^{2\pi }d\theta =\dfrac{4\pi }{3} \end{array} \]
953
The value of the triple integral \(\dfrac{4\pi }{3}\) equals the volume \(V\) of the solid. That is, \[ V=\dfrac{4\pi }{3} \hbox{ cubic units} \]