Double Iterated Integral question?
Use double iterated integrals to find the volume of the solid bounded by z=4-x^2-y^2 and z=0. I know the answer is 8*pi and I know that its shape is a dome, but I don’t know how to find the bounds. Thanks for any help
I knew it could be done with a triple integral but is there a way to do it with only a double integral? thanks for your help though!
The projection of the “dome” onto the xy plane is x^2 + y^2 ≤ 4, a circular region.
So, let’s try cylindrical coordinates. Then, this region is described by
0 ≤ z ≤ 4 – r^2, 0 ≤ r ≤ 2, and 0 ≤ θ ≤ 2π.
Therefore, the volume equals
∫∫∫ 1 dV
= ∫(θ = 0 to 2π) ∫(r = 0 to 2) ∫(z = 0 to 4 – r^2) 1 (r dz dr dθ)
= 2π ∫(r = 0 to 2) r(4 – r^2) dr
= 2π ∫(r = 0 to 2) (4r – r^3) dr
= 2π (2r^2 – r^4/4) {for r = 0 to 2}
= 8π.
I hope this helps!
Solid A Boys:FreeStyle off the top of the Dome!