
Jacob K. answered 12/16/21
McGill Grad for Nighttime Math Tutoring and Emergency Help
So, just to start, we know that this is a sphere with radius of 5. This is because 25=52, and just as in the equation of a circle in two dimensions, when x2+y2+z2≤something, it's usually a nice radius like this because we know its square root easily and it helps keep numbers nice. I'm just writing this to help you star to identify the patterns you can start to see in problems like this, as well as to help you visualize what exactly you're doing here. You're finding the volume of a sphere, something we know how to do well. You're just doing it with spherical coordinates now.
So, just to tidy ourselves here, I like to start by writing the equivalencies at the top when converting coordinates
In spherical coordinates, x2+y2+z2=ρ2
Further, the ball then becomes E={(ρ,θ,φ)∈R3¦ρ≤5, 0≤θ≤2π, 0≤φ≤π}. There is no magic behind that formula right there. Those are the bounds, unless otherwise specificed, when integrating spherical coordinates.
Similarly, as a constant, the volume dV becomes ρ2sin(φ)dρdφdθ. This is always true.
So, now we can convert
∫∫∫E(x2+y2+z2)dV =∫0π ∫02π ∫05ρρ2(ρ2sin(φ))dρdθdφ
I only switched dθ and dφ in order of integration for ease. It is the same as other multivariable integration. Further, notice that this is the same thing as (x2+y2+z2) * dV, as in the first integral. All we did was swap out the equivalencies we stated above for the sum of x/y/z and for dV, and we added bounds of integration because we know them for these terms.
Now, you just integrate like usual.
From inside to outside, ∫05ρ4sin(φ)dρ=ρ5sin(φ)/5, evaluated from 5 to 0, which is (55sin(φ))/5 - (0^5sin(φ))/5 = 625sin(φ).
Next, ∫02π625sin(φ)dθ=625sin(φ)*θ, evaluated from 2π to 0, which is just 2π*625sin(φ)=1250πsin(φ)
Lastly, integrate ∫0π1250πsin(φ)dφ=-1250πcos(φ) evaluated from π to 0. This is 2500π.
I hope this makes sense! Good luck