Question

..

Answer 1

Ok, when you draw them out, you see a region that is bounded on all sides. Correct?

If you could, please describe (or better yet, draw) the bounded region, so I can make sure we're on the same page.

Answer 2

Sorry for the sideways graph, @Vandreigan.

Answer 3

Looks good. We'll deal with the x integrals first.

We can see from the graph That the furthest right or left the area extends is limited by where f(x) = h(x) on the left, and where g(x) = h(x) on the right. These are the bounds on your integrals. Can you find out what x equals at these points?

Answer 4

is it (-2,2) for the point on the left? And then (4,2) on the right? Or am I supposed to do some sort of calculation to find that out?

Answer 5

What you have is correct. We only need the x values for now (since we'll be integrating over x.

Ok, the next step is to identify which functions are your upper and lower bounds for the area.

Answer 6

That was confusing to me because the examples I've had made it more obvious which ones were on top and which one was on bottom. I'm not sure with this one

Answer 7

For this particular area, we'll have to split it into two sections, as the lower bound changes in the middle of the shape.

Answer 8

Will we split it at the middle part where (0,0) is?

Answer 9

Look at the picture you drew.