# Thread: Help with setting up bonds in a triple integral

1. ## Help with setting up bonds in a triple integral

Hi,

Here is the problem i am working on...
evaluate the triple integral ∫ xy dV where E is the solid tetrahedron with vertices (0,0,0), (6,0,0), (0,1,0) and (0,0,7).

I sketched the region and got the following limits..
for x--- x=0 to x=6(1-y)
for y- - - y=0 to y=1
for z----z= 0 to z= 7(1-x/6-y)

i evaluated the integral with the order of integration dzdxdy and i got 15 but it's wrong.

can someone explain me if my bouds or order of integtation are wrong,
thank you

2. Originally Posted by Candy21
Hi,

Here is the problem i am working on...
evaluate the triple integral ∫ xy dV where E is the solid tetrahedron with vertices (0,0,0), (6,0,0), (0,1,0) and (0,0,7).

I sketched the region and got the following limits..
for x--- x=0 to x=6(1-y)
for y- - - y=0 to y=1
for z----z= 0 to z= 7(1-x/6-y)

i evaluated the integral with the order of integration dzdxdy (Everything is correct up to here!) and i got 15 but it's wrong.

can someone explain me if my bounds or order of integration are wrong,
thank you
The integral is $\displaystyle \int_0^1\!\!\!\int_0^{6(1-y)}\!\!\!\int_0^{7(1-\frac x6 - y)}\!\!\!xy\,dzdxdy = \int_0^1\!\!\!\int_0^{6(1-y)}\!\!\!7\bigl(1-\tfrac x6 - y\bigr)xy\,dxdy$. After doing the x-integral, I get $\displaystyle \int_0^1\!\!\!42y(1-y)^3\,dy = \frac{21}{10}$.