Question

Use triple integral to find the volume of region that is below z=8-x^2-y^2 above z=-√4x^2+4y^2 and inside x^2+y^2=4

Answer 1

Then,

$$V=\iiint\limits_{M}1dxdydz=\iint\limits_{x^2+y^2\le4}\left(\int\limits_{-2\sqrt{x^2+y^2}}^{8-\left(x^2+y^2\right)} 1dz\right)dxdy=\\[0.3cm] =\iint\limits_{x^2+y^2\le4}\left(8-\left(x^2+y^2\right)-\left(-2\sqrt{x^2+y^2}\right)\right)dxdy=\\[0.3cm] =\iint\limits_{x^2+y^2\le4}\left(8-\left(x^2+y^2\right)+2\sqrt{x^2+y^2}\right)dxdy=\\[0.3cm] =\left[\text{Pass to polar coordinates} :\\ \begin{array}{l} x=\rho\cdot\cos\theta\\ y=\rho\cdot\sin\theta\\ x^2+y^2=\rho^2\\ dxdy=\rho\cdot d\rho d\theta\\ \rho\in[0;2]\\ \theta\in[0;2\pi] \end{array}\right]=\\[0.3cm] =\int\limits_0^{2\pi}d\theta\cdot\left(\int\limits_0^2\left(8-\rho^2+2\rho\right)\rho d\rho\right)=\\[0.3cm] =\left.\theta\right|_0^{2\pi}\cdot\left(\left.\left(8\rho-\frac{\rho^3}{3}+\rho^2\right)\right|_0^2\right)=\\[0.3cm] =2\pi\cdot\left(8\cdot 2-\frac{2^3}{3}+2^2\right)=2\pi\cdot\left(\frac{16\cdot 3-8+4\cdot 3}{3}\right)=\\[0.3cm] =\frac{2\pi\cdot 52}{3}=\frac{104\pi}{3}\approx 108.9085$$

Conclusion,

$$\boxed{V=\frac{104\pi}{3}\approx108.9085}$$