24. find the volume of the solid that lies within both the cylinder $x^{2}+y^{2}=1$ and the sphere…

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 equation of the cylinder $x^{2}+y^{2}=1$ becomes $r = 1$, and the equation of the sphere $x^{2}+y^{2}+z^{2}=4$ becomes $r^{2}+z^{2}=4$, so $z=\pm\sqrt{4 - r^{2}}$.

Step2: Determine the limits of integration

For the cylinder $r$ ranges from $0$ to $1$, $\theta$ ranges from $0$ to $2\pi$, and $z$ ranges from $-\sqrt{4 - r^{2}}$ to $\sqrt{4 - r^{2}}$.

Step3: Set up the triple - integral for volume

The volume formula in cylindrical coordinates is $V=\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}\int_{z_1}^{z_2}r\ dz\ dr\ d\theta$. So $V=\int_{0}^{2\pi}\int_{0}^{1}\int_{-\sqrt{4 - r^{2}}}^{\sqrt{4 - r^{2}}}r\ dz\ dr\ d\theta$.

Step4: Integrate with respect to $z$ first

$\int_{0}^{2\pi}\int_{0}^{1}r\left[z\right]{-\sqrt{4 - r^{2}}}^{\sqrt{4 - r^{2}}}dr\ d\theta=\int{0}^{2\pi}\int_{0}^{1}r\left(\sqrt{4 - r^{2}}-\left(-\sqrt{4 - r^{2}}\right)\right)dr\ d\theta=\int_{0}^{2\pi}\int_{0}^{1}2r\sqrt{4 - r^{2}}dr\ d\theta$.

Step5: Use substitution for the integral with respect to $r$

Let $u = 4 - r^{2}$, then $du=-2r\ dr$. When $r = 0$, $u = 4$; when $r = 1$, $u = 3$. The integral becomes $\int_{0}^{2\pi}\int_{4}^{3}-\sqrt{u}du\ d\theta=\int_{0}^{2\pi}\int_{3}^{4}\sqrt{u}du\ d\theta$.

Step6: Integrate with respect to $u$

$\int_{0}^{2\pi}\left[\frac{2}{3}u^{\frac{3}{2}}\right]{3}^{4}d\theta=\int{0}^{2\pi}\frac{2}{3}(8 - 3\sqrt{3})d\theta$.

Step7: Integrate with respect to $\theta$

$\frac{2}{3}(8 - 3\sqrt{3})\int_{0}^{2\pi}d\theta=\frac{4\pi}{3}(8 - 3\sqrt{3})$.

Answer:

$\frac{4\pi}{3}(8 - 3\sqrt{3})$