Multiple Integrals

[Daniel_Feldman]

I need to find the volume of the bounded region in the first octant bounded by the surface z=1+(x-3)^2+3y^2 and the planes x=4 and y=2.


I need help setting up the integral. If i integrate with respect to dxdydz, my outer limits would be 0 to 10 I think (b/c the z-intercept is 10). I'm not sure how to find the inner limits, though. Any help would be much appreciated.

 

[mark07]

Easier way to approach these problems is to set it up as dz dx dy or dz dy dx, if possible. Because then you can visualize the xy-plane as the ground, visualize the solid above the ground, and find equations of lower and upper surfaces bounding the solid. In this case lower surface is z=0, upper is given in the problem.

After that, look at the solid from top and find the region R (on xy) which is the projection of your solid on xy-plane.

Then the setup would be

\(\displaystyle \L \iint_R \left( \int_0^{1+(x-3)^2+3y^2} dz \right) dA\)

To figure out the limits for R, try to visualize the projection of the solid on xy. You are cutting the paraboloid \(\displaystyle z=1+(x-3)^2+3y^2\) with the planes x=4 and y=2. Do you see that R is simply the following rectangle?

\(\displaystyle \L 0 \leq x \leq 4 \text{ \; and \; } 0\leq y \leq 2\)

I think you can finish now...

 

[Daniel_Feldman]

Ok, thanks, so the limits on x are just 0 to 4 and the limits on y are 0 to 2, and then I solve.