a triple integral question?

Question

I just want the know the boundary points. And if u know, tell me how u get it. The question is: Consider z= (x^2+y^2)^1/2 and z=a, where a is constant and positive. Define W to be teh solid below z=a and above z=(x^2+y^2)^1/2, and define the mass-density of W and the rest is that u have to integrate k(x^2+y^2+z^2)^1/2 but i can do that from there. ANyways i suspect the theta will be 0 to 2Pi so i need to know what is rho and phi...
it is spherical coordiantes btw....

Answer 1

x = r cos φ cos θ
y = r cos φ sin θ
z = r sin φ

We're clearly in positive z territory, so we know φ is limited to within [0, π/2]; and it's symmetrical in x and y, so we will have θ going across the whole range [0, 2π], and we should expect θ to disappear from the integrand.

We want (x^2 + y^2)^(1/2) <= z <= a
<=> √(r^2 cos^2 φ (cos^2 θ + sin^2 θ)) <= r sin φ <= a
<=> r cos φ <= r sin φ <= a
<=> cos φ <= sin φ and 0 <= r <= a/sin φ.
<=> φ ∈ [π/4, π/2] and r ∈ [0, a / sin φ].

So the integral is
∫(π/4 to π/2) ∫(0 to a/sin φ) ∫(0 to 2π) kr (r^2 cos φ) dθ dr dφ
= ∫(π/4 to π/2) ∫(0 to a/sin φ) 2πk r^3 cos φ dr dφ
= ∫(π/4 to π/2) 2πk cos φ (a / sin φ)^4 / 4 dφ
Let u = sin φ, du = cos φ dφ
= ∫(1/√2 to 1) 2πk (a^4/4) du / u^4
= 2πk (a^4/4) [u^(-3) / (-3) ] [1/√2 to 1]
= πk (a^4/2) [-1/3 + 2√2 / 3]
= πka^4 (2√2 - 1) / 6.