Use a triple integral to find volume of a solid..

[hank]

I just need to know if I've set this problem up correctly, and I'd like it if someone can verify my answer:

The solid bounded by the surface y = x^2 and the planes y + z = 4 and z = 0.

Setup:

\(\displaystyle 2\int^{2}_{0}\int^{4-x^2}_{0}\int^{4 - y}_{0} dzdydx\)

The answer I get is \(\displaystyle \frac{256}{10}\).
The answer in the book is \(\displaystyle \frac{256}{15}\).

 

[galactus]

Try this:

\(\displaystyle 2\int_{0}^{2}\int_{-x^{2}}^{x^{2}}\int_{0}^{4-x^{2}}dzdydx=\frac{256}{15}\)

 

[hank]

Try this:

\(\displaystyle 2\int_{0}^{2}\int_{-x^{2}}^{x^{2}}\int_{0}^{4-x^{2}}dzdydx=\frac{256}{15}\)

I got it, thanks.