Integral Calculus

**Abu-Khalil** · May 3rd 2009, 04:30 PM

Find the volume between $x+2y+z=4,z=x+y,x=0,y=0$ .

I got

, but it gives me a negative volume

**NonCommAlg** · May 3rd 2009, 04:47 PM

Originally Posted by Abu-Khalil

Find the volume between $x+2y+z=4,z=x+y,x=0,y=0$ .

I got

, but it gives me a negative volume

the correct one is this: $\int_0^2 \int_0^{\frac{4 - 2x}{3}} \int_{x+y}^{4-x-2y} dz dy dx.$

**TheEmptySet** · May 3rd 2009, 04:48 PM

Originally Posted by Abu-Khalil

Find the volume between $x+2y+z=4,z=x+y,x=0,y=0$ .

I got

, but it gives me a negative volume

I think you have a mistake in your limits of integration

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

and x goes from 0 to 2.

**Abu-Khalil** · May 3rd 2009, 05:02 PM

Oh Ok, i got it. So.. i must replace the $z$ from one equation to other, right?

Thread: Integral Calculus