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.)

Accepted Solution

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

The radius is a constant of [tex]\sqrt{5}[/tex] 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 [tex]z=2[/tex] plane which means that it forms a triangle of hypotenuse [tex]\sqrt{5}[/tex] and an opposite leg of [tex]2[/tex]. This produces an angle of [tex]\phi = 1.107[/tex] so our limits for [tex]0 \leq \phi \leq 1.107[/tex]

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

[tex]\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{1.107}\int_{0}^{\sqrt{5}} d\rho d\phi d\theta[/tex]