# Triple Intergral Bounds Help

• NZer

#### NZer

Hi guys,
Im currently revising for my exams and I encountered a problem I hope someone will be able to help me with.

## Homework Statement

Find the volume of the region of space bounded by:
The planes x=0, y=0, z=0, z=3-2x+y and the surface y=1-x^2

## Homework Equations

$$\int \int \int _R 1\, dV$$

## The Attempt at a Solution

First I decided to integrate with respect to the z direction as I wouldn't have to worry about splitting up the region yet.

$$\int \int \int _0 ^{3-2x+y} 1\, dz\, dy\, dx$$

$$= \int \int 3-2x+y\, dy \, dx$$

ok. But now I have a problem due to the surface y=1-x^2 cutting our region defined by the 4 planes. Can we split the region and choose our bounds like below?

$$= \int _1 ^{3/2} \int _{1-x^2} ^{2x-3} 3-2x+y\, dy\, dx \; + \int _0 ^1 \int _0 ^{2x-3} 3-2x+y\, dy \, dx$$

Thanks.

i have something simpler:

say f(x,y) = 3-2x+y

so now volume is

int (0,1) . int (0, 1-x^2) f(x,y) dy.dx

If you can picture the region, the volume you want appears to be confined to the 2nd octant. For this, you'll want the limits of integration for y to be from 0 to 1-x^2. Why is any "splitting of region" necessary?

Thanks for the reply rootX, Defennder

Indeed you are both right.
After scratching my head for a while I noticed I drew my diagram slightly wrong (I had the plane as z=3+2x-y lol) so my region projected onto the xy-plane was piece-wise defined.

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