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In this problem there is an oblique plane and they are trying to find the volume ABOVE the plane.

I understand how bottom and top limit of this question was found and it also makes sense to me how upper limit of integration was found when integrating with respect to x after it is integrated by Z but I dont get why when integrated with respect to x's lower limit is 0 because isn't the oblique plane changing the lower bound of x?

I am getting really frustrated and hazy when trying to figure out the upper and lower limits for triple integrals with variables for limits....isn't there an easier way to visualize or even come up with these limits?

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I think your confusion lies in your statement "This equation is the one being integrated". From your image, it appears that it's the density $\rho(x,y,z)=x^2yz$ that's being integrated, and the equation $2x+3z+y=6$ is the equation of the oblique plane in the image that forms the lower boundary of the volume integration. Thus the answer to your question is basically "by definition": The lower limit is given by solving that equation for $z$ simply because that equation is meant to specify the boundary of the volume of integration.

By the way, there's something wrong in the last line, where you have one integral sign and two differential symbols, but you should have three integral signs and three differential symbols.

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OOOOOOO sorry it was a math video I was watching after doing an entire day of integration and I didn't pay enough attention to the part where he said volume ABOVE the oblique plane. Thanks to you it all makes sense now! –  Raynos Dec 5 '12 at 22:17

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