Question

Write a triple integral including limits of integration that gives the volume of the cap of the solid sphere x2+y2+z2≤5 cut off by the plane z=2 and restricted to the first octant. (in your integral, use theta, rho, and phi for θ, ρ and ϕ, as needed.)

Answer 1

Since it is in the first octant, if we let $\theta$ be the angle formed in the xy plane, then its limits are simply $0 \leq \theta \leq \frac{\pi}{2}$.

The radius is a constant of $\sqrt{5}$ so we integrate from [tex0 \leq \rho \leq \sqrt 5[/tex]

The yz/xz plane doesn't go the full 90 degrees. Instead, it goes to the $z=2$ plane which means that it forms a triangle of hypotenuse $\sqrt{5}$ and an opposite leg of $2$. This produces an angle of $\phi = 1.107$ so our limits for $0 \leq \phi \leq 1.107$

We are just integrating a constant of 1, then we get:

$\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{1.107}\int_{0}^{\sqrt{5}} d\rho d\phi d\theta$