Question

Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder X^2+Y^2=4 and the plane X+Z=4.

Answer 1

Answer:

$4\pi- \frac{8}{3}$

Step-by-step explanation:

The solid that you described is shown in the image of below. To compute the volume we use triple integrals. Observe that

$0\leq x\leq 2$

$0\leq y \leq \sqrt{4-x^2}$

$0\leq z \leq 4-x$

So, the volumen of the solid is given by

$\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}\int_{0}^{4-x}1\, dzdydx=\int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}4-x\, dydx=\int_{0}^{2}(4-x)\sqrt{4-x^2}\,dx$

This last integral equals $4\pi -\frac{8}{3}$