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), (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 (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 formula for mass (m=\iiint_{V}\rho(x,y,z)dV). The limits of integration are: (0\leq\theta\leq2\pi), (0\leq r\leq1), (-\sqrt{4 - r^{2}}\leq z\leq\sqrt{4 - r^{2}}). So the integral is (m=\int_{0}^{2\pi}\int_{0}^{1}\int_{-\sqrt{4 - r^{2}}}^{\sqrt{4 - r^{2}}}r^{2}\cos^{2}\theta\cdot r\ dz\ dr\ d\theta).

Step3: Integrate with respect to (z)

(\int_{-\sqrt{4 - r^{2}}}^{\sqrt{4 - r^{2}}}dz=\left[z\right]{-\sqrt{4 - r^{2}}}^{\sqrt{4 - r^{2}}}=2\sqrt{4 - r^{2}}). The integral becomes (m = 2\int{0}^{2\pi}\cos^{2}\theta\ d\theta\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr).

Step4: Use the double - angle formula for (\cos^{2}\theta)

We know that (\cos^{2}\theta=\frac{1 + \cos(2\theta)}{2}). Then (\int_{0}^{2\pi}\cos^{2}\theta\ d\theta=\int_{0}^{2\pi}\frac{1+\cos(2\theta)}{2}d\theta=\left[\frac{\theta}{2}+\frac{\sin(2\theta)}{4}\right]_{0}^{2\pi}=\pi).

Step5: Use substitution for the integral in (r)

Let (u = 4 - r^{2}), then (du=-2r\ dr) and (r^{2}=4 - u). When (r = 0), (u = 4); when (r = 1), (u = 3). (\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr=-\frac{1}{2}\int_{4}^{3}(4 - u)\sqrt{u}du=\frac{1}{2}\int_{3}^{4}(4u^{\frac{1}{2}}-u^{\frac{3}{2}})du). [ \begin{align*} \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{40\sqrt{3}-18\sqrt{3}}{5}\right)\ &=\frac{1}{2}\left(\frac{128}{15}-\frac{22\sqrt{3}}{5}\right) \end{align*} ]

Step6: Calculate the final mass

Since (m = 2\pi\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr), substituting the value of (\int_{0}^{1}r^{3}\sqrt{4 - r^{2}}dr) we found above. [ \begin{align*} m&=2\pi\times\frac{1}{2}\left(\frac{128}{15}-\frac{22\sqrt{3}}{5}\right)\ &=\pi\left(\frac{128 - 66\sqrt{3}}{15}\right) \end{align*} ]

Answer:

(m=\frac{\pi(128 - 66\sqrt{3})}{15})