# triple integral volume problem

• Jun 6th 2009, 08:15 PM
Frostking
triple integral volume problem
I need to find the volume of region bound by z = x + y z = 10 and the planes x= 0 and y = 0

I set up an integral from 0 to 10 then from 0 to 10 - x and from 10 to x + y dzdydx with the function being one dzdydx

I get an answer 5 times the amount I should. Could someone please tell me if I miscalculated the bounds on one of my integrals??? Thanks so much for the assistance, Frostking
• Jun 7th 2009, 05:07 AM
Scott H
Here, $z=x+y$ is a plane sloping diagonally upward along the $x$- and $y$-axes that forms a tetrahedron with the planes $x,\,y=0$ and $z=10$. We can always use the formula for the volume of a cone:

$V=\frac{1}{3}Bh,$

where $B$ is the base area, here $50$. You are correct that the integral to evaluate is

$\int_0^{10}\int_0^{10\,-\,x}\int_{x\,+\,y}^{10}dz\,dy\,dx.$

Carrying out two steps, we find that this equals

\begin{aligned}
\int_0^{10}\int_0^{10\,-\,x}(10-x-y)\,dy\,dx &=\int_0^{10}\left[10y-xy-\frac{y^2}{2}\right]_0^{10\,-\,x}\,dx\\
&=\int_0^{10}\left(10(10-x)-x(10-x)-\frac{1}{2}(10-x)^2\right)\,dx\\
&=\int_0^{10}\left(100-10x-10x+x^2-\frac{x^2}{2}+10x-50\right)\,dx.
\end{aligned}