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## Homework Statement

Find the volume of the region in the 1st octant bounded above by the surface z=4-x^2-y and below by the plane z=3.

Answer = 4/15

## Homework Equations

V = [tex]\int\int\int dV[/tex]

## The Attempt at a Solution

I'm having trouble determining the upper and lower z limits.

I partly don't understand what to do, is the function z=3 or z=4-x^2-y that I am integrating.

[tex] \int_0^2 \int_0^{4-x^{2}} \int_0^3 \left (1) \right dz.dy.dx [/tex]

skipping the tedious integration working out.

V=16 which is not the right answer.

I just realised my diagram is wrong, as I drew it as a parabola concave down in the wrong plane.

I've written the new integral as

[tex] \int_0^2 \int_0^{4-x^{2}} \int_0^{3} \left (4-x^2-y) \right dz.dy.dx [/tex]

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