Find the volume $V$ common to the two cylinders $x^{2}+y^{2}=1$ and $x^{2}+z^{2}=1$.

Solution Each cylinder has radius $1$; their axes are perpendicular to each other and lie on the $z$-axis and $y$-axis, respectively. Figure 22(a) shows the portion of the solid lying in the first octant. If we find the volume of this portion of the solid, then since the volume in each octant is the same, the volume $V$ is eight times the volume in the first octant. That is, if $z=f(x,y) =\sqrt{1-x^{2}},$ then $$V=8\displaystyle\iint\limits_{\kern-3ptR}f(x,y) \,{\it dA}=8\displaystyle\iint\limits_{\kern-3ptR}(1-x^{2})^{1/2}\, {\it dA}$$

where $R$ is the region in the first quadrant inside the circle $x^{2}+y^{2}=1.$ See Figure 22(b).

Consider the integrand. It is simpler to integrate with respect to $y$ first. (Do you see why?) So, we treat $R$ as an $x$-simple region. (If this approach fails, we would try treating $R$ as a $y$-simple region.) If $R$ is $x$-simple, then $y$ varies from $0$ to $(1-x^{2})^{1/2}$, $0\leq x\leq 1$. The volume $V$ is $$\begin{eqnarray*} V &=&8\displaystyle\iint\limits_{\kern-3ptR}(1-x^{2})^{1/2}\, {\it dy}\,{\it dx}=8\int_{0}^{1}\left[ \int_{0}^{ \sqrt{1-x^{2}}}(1-x^{2})^{1/2}\,{\it dy}\right] \! {\it dx}\\[5pt] &=&8\int_{0}^{1}\Big[ (1-x^{2})^{1/2}y\Big] _{0}^{\sqrt{1-x^{2}}}\,{\it dx} \notag \\[5pt] &=&8\int_{0}^{1}(1-x^{2})\,{\it dx}=8\left[ x-\dfrac{x^{3}}{3}\right] _{0}^{1}= \dfrac{16}{3}\hbox{ cubic units} \end{eqnarray*}$$