# Integral Calculus

• May 3rd 2009, 04:30 PM
Abu-Khalil
Integral Calculus
Find the volume between $\displaystyle x+2y+z=4,z=x+y,x=0,y=0$.

I got http://www.fmat.cl/tex/d99a6824adf2f...9c291cbfda.png, but it gives me a negative volume (Punch)
• May 3rd 2009, 04:47 PM
NonCommAlg
Quote:

Originally Posted by Abu-Khalil
Find the volume between $\displaystyle x+2y+z=4,z=x+y,x=0,y=0$.

I got http://www.fmat.cl/tex/d99a6824adf2f...9c291cbfda.png, but it gives me a negative volume (Punch)

the correct one is this: $\displaystyle \int_0^2 \int_0^{\frac{4 - 2x}{3}} \int_{x+y}^{4-x-2y} dz dy dx.$
• May 3rd 2009, 04:48 PM
TheEmptySet
Quote:

Originally Posted by Abu-Khalil
Find the volume between $\displaystyle x+2y+z=4,z=x+y,x=0,y=0$.

I got http://www.fmat.cl/tex/d99a6824adf2f...9c291cbfda.png, but it gives me a negative volume (Punch)

I think you have a mistake in your limits of integration

$\displaystyle x+2y+x+y=4 \iff 3y=-2x+4 \iff y =-\frac{2}{3}+\frac{4}{3}$

and x goes from 0 to 2.
• May 3rd 2009, 05:02 PM
Abu-Khalil
Oh Ok, i got it. So.. i must replace the $\displaystyle z$ from one equation to other, right?