Sketch the region R whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area.
Step 11 of 3
Step 11 of 13
The problem defines two integration regions (since we have two iterated integrals).
We determine the first region of integration from the first integral, which can be written as:
Since under the integral first goes and then , the limits of the internal integral are over the variable .
That is, takes values between:
The limits of the external integral are the -axis boundaries. Therefore can take values between: