# Triple Integral, already solved, need checked

1. Dec 16, 2009

### King Tony

1. The problem statement, all variables and given/known data
Let S be the region in the first octant under the plane 3x + 2y +z = 4. Find the volume of S.

2. Relevant equations
idk?

3. The attempt at a solution

$$\int^{\frac{4}{3}}_{0}\int^{\frac{3}{2}x + 2}_{0}\int^{-3x - 2y + 4}_{0}dzdydx$$

= $$\int^{\frac{4}{3}}_{0}\int^{\frac{3}{2}x + 2}_{0}(-3x - 2y + 4)dydx$$

= $$\int^{\frac{4}{3}}_{0}(-\frac{15}{2}x^{2} - 6x + 4)dx$$

= 448/27

Last edited: Dec 16, 2009
2. Dec 16, 2009

### King Tony

Sorry, took a little while to figure out the symbols and dealies, pretty sure it's good to go right now.

3. Dec 16, 2009

### arildno

Mumble, mumble..gotta check the bounds:

The upper line in the z=0 plane obeys the equation 3x+2y=4

Thus,
$$0\leq{x}\leq\frac{4}{3}, 0\leq{y}\leq{2}-\frac{3}{2}x$$
Thus, you've got a sign wrong in the y-bound, according to my view.

4. Dec 16, 2009

### King Tony

You're so right, thankyou!