Quiz 9, Complete Solution for Problem 1

We want to set up the integral of f over the solid. We will follow the procedure in:

How to find the limits of integration, given a solid.

1.) Draw the solid S.

Here's a picture of the solid.

2.) Find equations for each of the smooth pieces of the boundary of R.

This is easy; the boundary has two smooth pieces, and the equations are z= 25-x²-y², and z=-25+x²+y².

3.) Decide which variable you want to use for the first integration.

We decide to integrate with respect to z first, because if we draw a bunch of vertical lines, the ones that intersect our solid all enter by the lower boundary and exit by the top.

4.) Look at where the vertical segments enter the solid S. Use the equations you found in step 2 to write a formula for the z value where the segments first enter the solid in terms of x and y. This is the lower limit of the z integral.

This is easy; we see that the lower limit is just -25+x²+y² because this is the formula for the z-coordinates of points on the lower boundary. The upper limit must be 25-x²-y², so we can skip to step 6

6.) Draw the shadow of the solid.

Here is a picture of the shadow:

The shadow is the disk of radius 5 centered at the origin. How did we get this? A point in the xy-plane is in the shadow if the vertical line running through the point intersects the solid. Looking at the picture of the solid, it's clear that the edge of the shadow will be the set where the top and bottom of the solid intersect. The points that are on both the top and bottom of the solid, satisfy both equations z= 25-x²-y², z=-25+x²+y².

If we eliminate z we get 25-x²-y²=-25+x²+y². Combining terms we get 25=x²+y², which is the equation of a circle of radius 5.

7.) Find the limits of integration for the integral over the shadow.

Here is a picture of the shadow, with horizontal lines drawn through it. We see that the lines all enter by the left side of the circle and exit by the right. On the left side of the circle we know that x=-(25-y²)^1/2, so the lower limit of the x integral is -(25-y²)^1/2. Meanwhile, on the right side of the circle, x=(25-y²)^1/2, so the upper limit of the x integral is (25-y²)^1/2. Since the horizontal lines first hit the circle at y=-5 and last hit the circle at y=5, we know that the limits for the y-integral are -5 and 5. So the final answer is:

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