Find the volume of the region D bounded by the hemisphere $y=\sqrt{4-x^2-z^2}$ and the planes $y=x\ $,$\ y= \sqrt3x$ by using polar coordinates.
Working:
I have calculated $x^2+y^2+z^2=4$ so am I correct in stating $0\leq r\leq 2$? But I am really unsure of what to do next! I am relatively new to triple integration with spherical coordinates and would appreciate any help!
So you found $0\leq r \leq 2$. Think about the solid you're integrating, it is half of a sphere in the positive $y$ region of radius $2$ bounded by the planes $y=x$ and $y= \sqrt 3x$, you will get $0\leq\phi\leq \pi$ and $\frac \pi 4 \leq \theta \leq \frac \pi 3$. $\theta$ is the tricky one to find in this case, try visualizing it by graphing $y=x$ and $y= \sqrt 3x$ on the $xy-$plane (you end up with those special triangles you had to memorize in high school). Remember $\theta$ is the same angle counter-clockwise from the positive $x$-axis and $\phi$ is the angle clockwise from the positive $z$-axis. The area would be: $$\int \int \int_D1dV=\int_0^2\int_0^\pi\int_{\frac \pi4}^{\frac \pi3}r^2\sin(\phi)d\theta d\phi dr$$ In this case you're integrating over a rectangular region(in spherical coords) so it doesn't matter which order you integrate. I would recommend graphing a whole bunch of these $3D$ functions on wolframalpha as being able to visualize them helps a TON, Good luck!