Calculating the volume of a solid in the first octant.

Question

I am asked to verify the divergence theorem for a vector field in the region of the first octant limited by $x=2$ and $y^2+z^2=9$, so I need to calculate the volume of this solid. However I don't know how to delimitate the bounds of the triple integral, of course $0\leq x \leq 2$, $-\sqrt{9-z^2}\leq y\leq \sqrt{9-z^2}$ but I don't know how to follow.

Answer 1

As Ted Shifrin said this solid is a quarter of a cylinder. $y^2+ z^2= 9$ is the cylinder with axis along the x-axis and radius 3. Requiring that x= 2 and x= 0 (since this is in the first octant) gives a cylinder with volume $\pi(3^2)(2)= 18\pi$. Since this is in the first quadrant, the volume is $\frac{18\pi}{4}= \frac{9}{2}\pi$.

To do the volume in cylindrical coordinates, I would swap the x and z coordinates: The region is that portion of the cylinder $x^2+ y^2= 9$ in the first quadrant from z= 0 to z= 2. That would be $\int_0^2\int_0^{\pi/2}\int_0^3 rdrd\theta dz$.