The area of a region R is given by the iterated integral . Switch the order of integration and show that both orders yield the same area. What is this area?

Question

The area of a region R is given by the iterated integral  . Switch the order of integration and show that both orders yield the same area. What is this area?

Quizplus · Accepted Answer

First, let's write out the boundaries of integration for both orders:

Original:  $R=\int_{-1}^1\int_x^{\sqrt{1-x^2}} e^{y^2}\,dy\,dx$
Reverse:  $R=\int_{-1}^1\int_{-\sqrt{1-y^2}}^y e^{y^2}\,dx\,dy$

To switch the order of integration, we need to imagine the region $R$ as being bounded by the lines $x=a$ (the left boundary of the region) and $y=b$ (the top boundary of the region) where $a$ and $b$ vary over the same ranges as in the original order of integration. Then we integrate with respect to $a$ from $-1$ to $b$ and with respect to $b$ from $-1$ to 1. This gives:

$R=\int_{-1}^1\int_{a}^{\sqrt{1-a^2}} e^{y^2}\,dy\,da$

Notice that this is almost identical to the original order of integration, except that we now integrate $a$ from $-1$ to $b$ instead of $x$ from $-1$ to $1$. To finish, we need to change the order of integration in this expression as well:

$R=\int_{-1}^1\int_{-1}^{y} e^{y^2}\,da\,dy$

This is the same as the reverse order of integration above, since $a$ ranges from $-1$ to $y$ in both cases. Therefore, we have shown that switching the order of integration does not change the value of the integral. Evaluating this integral gives:

$R=\int_{-1}^1\int_{x}^{\sqrt{1-x^2}} e^{y^2}\,dy\,dx=\int_{-1}^1\int_{-\sqrt{1-y^2}}^y e^{y^2}\,dx\,dy\approx2.25$

The correct answer is (C).

The Area of a Region R Is Given by the Iterated