Find the volume of the region B bounded above by the sphere $x^2 + y^2
+ z^2 = a^2$ and below by the plane $z = b$, where $a > b > 0$.
Here I worked out $z=\sqrt{a^2-r^2}$ and $x^2+y^2=a^2-b^2$ so $0≤r ≤\sqrt{a^2-b^2}$ and $0≤θ≤2π$ and $\int_{0}^{2\pi }\int_{0}^{\sqrt{a^2-b^2}} r(a^2-r^2)^\frac{1}{2}drd\theta$ = $(0.5(a^2-b^2)a^2-0.25(a^2-b^2)^2)2π$ but the book has a different answer. I dont know where I have gone wrong i havent made any mistakes in the calculation.
Edit: I think they have used cylindrical coordinates with triple integral whereas I've used polar with doouble integral but i thought it doesnt make a difference?
Best Answer
First determine the volume you integrate over. The region in the $x,y$ plane is bounded by $x^2+y^2=a^2$. The height $z-$coordinate is bounded from above by $z=b$ and from below by $z=\sqrt{a^2-x^2-y^2}$. Now it is wise to switch to cylinder coordinates to calculate the final integral.
$$ V=\int_{0}^a\int_{0}^{2\pi}\int_{\sqrt{a^2-r^2}}^{b}r\,dzd\theta dr $$