an object occupies the space inside both the cylinder $x^{2}+y^{2}=1$ and the sphere $x^{2}+y^{2}+z^{2}=4$…

Explanation:

Step1: Convert to cylindrical coordinates

In cylindrical coordinates, (x = r\cos\theta), (y = r\sin\theta), (z = z), and (dV=r\ dz\ dr\ d\theta). The cylinder (x^{2}+y^{2}=1) becomes (r = 1), and the sphere (x^{2}+y^{2}+z^{2}=4) becomes (r^{2}+z^{2}=4) or (z=\pm\sqrt{4 - r^{2}}). The density function (\rho(x,y,z)=x^{2}) becomes (\rho(r,\theta,z)=r^{2}\cos^{2}\theta).

Step2: Set up the triple - integral for mass

The mass (m=\int\int\int_{V}\rho(x,y,z)dV). Using cylindrical coordinates and symmetry (since the region and the density function (x^{2}) have symmetry about the (z) - axis and the (xy) - plane), we can set up the integral as: [ \begin{align*} m&=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\int_{0}^{\sqrt{4 - r^{2}}}r^{2}\cos^{2}\theta\cdot r\ dz\ dr\ d\theta\ &=4\int_{0}^{\frac{\pi}{2}}\cos^{2}\theta\ d\theta\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr \end{align*} ] First, use the double - angle formula (\cos^{2}\theta=\frac{1 + \cos(2\theta)}{2}), then (\int_{0}^{\frac{\pi}{2}}\cos^{2}\theta\ d\theta=\int_{0}^{\frac{\pi}{2}}\frac{1+\cos(2\theta)}{2}d\theta=\left[\frac{\theta}{2}+\frac{\sin(2\theta)}{4}\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{4})

For (\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr), use substitution (u = 4 - r^{2}), then (r^{2}=4 - u) and (dr=-\frac{du}{2r}). When (r = 0), (u = 4); when (r = 1), (u = 3) [ \begin{align*} \int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr&=\frac{1}{2}\int_{3}^{4}(4 - u)\sqrt{u}du\ &=\frac{1}{2}\int_{3}^{4}(4u^{\frac{1}{2}}-u^{\frac{3}{2}})du\ &=\frac{1}{2}\left[4\times\frac{2}{3}u^{\frac{3}{2}}-\frac{2}{5}u^{\frac{5}{2}}\right]_{3}^{4}\ &=\frac{1}{2}\left(\frac{8}{3}\times4^{\frac{3}{2}}-\frac{2}{5}\times4^{\frac{5}{2}}-\frac{8}{3}\times3^{\frac{3}{2}}+\frac{2}{5}\times3^{\frac{5}{2}}\right)\ &=\frac{1}{2}\left(\frac{64}{3}-\frac{64}{5}-\frac{8\times3\sqrt{3}}{3}+\frac{2\times3^{2}\sqrt{3}}{5}\right)\ &=\frac{1}{2}\left(\frac{320 - 192}{15}-8\sqrt{3}+\frac{18\sqrt{3}}{5}\right)\ &=\frac{1}{2}\left(\frac{128}{15}-\frac{22\sqrt{3}}{5}\right) \end{align*} ]

Another way for (\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr): Let (t=\sqrt{4 - r^{2}}), (r^{2}=4 - t^{2}), (r\ dr=-t\ dt) [ \begin{align*} \int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr&=\int_{2}^{\sqrt{3}}(4 - t^{2})t(-t\ dt)\ &=\int_{\sqrt{3}}^{2}(4t^{2}-t^{4})dt\ &=\left[\frac{4t^{3}}{3}-\frac{t^{5}}{5}\right]_{\sqrt{3}}^{2}\ &=\frac{32}{3}-\frac{32}{5}-\left(4\sqrt{3}-\frac{9\sqrt{3}}{5}\right)\ &=\frac{64}{15}-\frac{11\sqrt{3}}{5} \end{align*} ]

Using symmetry and a better substitution: [ \begin{align*} m&=4\times\frac{\pi}{4}\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr\ \end{align*} ] Let (u = 4 - r^{2}), (r^{2}=4 - u), (r\ dr=-\frac{1}{2}du) [ \begin{align*} \int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr&=\frac{1}{2}\int_{3}^{4}(4 - u)\sqrt{u}du\ &=\frac{1}{2}\int_{3}^{4}(4u^{\frac{1}{2}}-u^{\frac{3}{2}})du\ &=\frac{1}{2}\left[\frac{8}{3}u^{\frac{3}{2}}-\frac{2}{5}u^{\frac{5}{2}}\right]_{3}^{4}\ &=\frac{1}{2}\left(\frac{8}{3}\times8-\frac{2}{5}\times32-\frac{8}{3}\times3\sqrt{3}+\frac{2}{5}\times9\sqrt{3}\right)\ &=\frac{1}{2}\left(\frac{64}{3}-\frac{64}{5}-8\sqrt{3}+\frac{18\sqrt{3}}{5}\right)\ &=\frac{1}{2}\left(\frac{320 - 192}{15}-\frac{40\sqrt{3}-18\sqrt{3}}{5}\right)\ &=\frac{1}{2}\left(\frac{128}{15}-\frac{22\sqrt{3}}{5}\right) \end{align*} ]

Using the substitution (v = r^{2}), (dv = 2r\ dr) [ \begin{align*} \int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr&=\frac{1}{2}\int_{0}^{1}v\sqrt{4 - v}dv \end{align*} ] Let (s = 4 - v), (v = 4 - s), (dv=-ds) [ \begin{align*} \frac{1}{2}\int_{0}^{1}v\sqrt{4 - v}dv&=\frac{1}{2}\int_{3}^{4}(4 - s)\sqrt{s}(-ds)\ &=\frac{1}{2}\int_{3}^{4}(4s^{\frac{1}{2}}-s^{\frac{3}{2}})ds\ &=\frac{1}{2}\left[\frac{8}{3}s^{\frac{3}{2}}-\frac{2}{5}s^{\frac{5}{2}}\right]_{3}^{4}\ &=\frac{1}{2}\left(\frac{64}{3}-\frac{64}{5}-\frac{8\times3\sqrt{3}}{3}+\frac{2\times9\sqrt{3}}{5}\right)\ &=\frac{1}{2}\left(\frac{320 - 192}{15}-\frac{40\sqrt{3}-18\sqrt{3}}{5}\right)\ &=\frac{1}{2}\left(\frac{128}{15}-\frac{22\sqrt{3}}{5}\right) \end{align*} ]

Using the formula (\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a + b - x)dx) is not very helpful here.

Using the fact that (\int r^{3}\sqrt{4 - r^{2}}dr): Let (w = 4 - r^{2}), (r^{2}=4 - w), (r\ dr=-\frac{1}{2}dw) [ \begin{align*} \int r^{3}\sqrt{4 - r^{2}}dr&=-\frac{1}{2}\int(4 - w)\sqrt{w}dw\ &=-\frac{1}{2}\int(4w^{\frac{1}{2}}-w^{\frac{3}{2}})dw\ &=-\frac{1}{2}\left(\frac{8}{3}w^{\frac{3}{2}}-\frac{2}{5}w^{\frac{5}{2}}\right)+C\ &=-\frac{4}{3}(4 - r^{2})^{\frac{3}{2}}+\frac{1}{5}(4 - r^{2})^{\frac{5}{2}}+C \end{align*} ] [ \begin{align*} \int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr&=\left[-\frac{4}{3}(4 - r^{2})^{\frac{3}{2}}+\frac{1}{5}(4 - r^{2})^{\frac{5}{2}}\right]_{0}^{1}\ &=-\frac{4}{3}(3)^{\frac{3}{2}}+\frac{1}{5}(3)^{\frac{5}{2}}+\frac{4}{3}\times8-\frac{1}{5}\times32\ &=\frac{32}{3}-\frac{32}{5}-4\sqrt{3}+\frac{9\sqrt{3}}{5}\ &=\frac{160 - 96}{15}-\frac{20\sqrt{3}-9\sqrt{3}}{5}\ &=\frac{64}{15}-\frac{11\sqrt{3}}{5} \end{align*} ]

[ \begin{align*} m&=4\times\frac{\pi}{4}\times\frac{11}{15}\ &=\frac{11\pi}{15} \end{align*} ]

Answer:

(\frac{11\pi}{15})