Computing the volume of a solid.?

Question

Compute the volume of the solid enclosed by the cylinders z = x^2, y = x^2, and the planes z = 0, y = 4. For the most part, I resolved the problem correctly. I found that x^2 < y < 4, -2 < x < 2, but I mistakenly arrived at 0 < z < y. Originally, I had z between 0 and x^2, but thought that since y = x^2, it would also be fair to say 0 < z < y. Why was this assumption wrong? I supposed I should have used a double integral instead of integrating 1, but I was having a very tough time telling what the surface looked like during the exam.

Thanks!

kb · Accepted Answer

When finding the volume under z = f(x,y) over some region in the xy-plane, you want to treat any z equations for the integrand (not mingling it any further with x and y).

(If you wanted to write x = f(y, z), then you need bounds in the yz-plane, ansd what you wrote could be useful.)
----------
Bounds for z:
z = 0 to z = x^2.

The region on the xy-plane is bounded between y = x^2 and y = 4.
==> y = x^2 to y = 4 with x in [-2, 2] (which is easy to sketch).

So, the volume equals
∫(x = -2 to 2) ∫(y = x^2 to 4) ∫(z = 0 to x^2) 1 dz dy dx
or ∫(x = -2 to 2) ∫(y = x^2 to 4) (x^2 - 0) dy dx.

Either way, this simplifies to
∫(x = -2 to 2) x^2(4 - x^2) dx
= 2 ∫(x = 0 to 2) (4x^2 - x^4) dx, since the integrand is even
= 2(4x^3/3 - x^5/5) {for x = 0 to 2}
= 128/15.

I hope this helps!

Trending News

Computing the volume of a solid.?

1 Answer