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### Section 4-3 : Double Integrals over General Regions

12. Use a double integral to determine the area of the region bounded by $$y = 1 - {x^2}$$ and $$y = {x^2} - 3$$.

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Okay, we know that the area of any region $$D$$ can be found by evaluating the following double integral.

$A = \iint\limits_{D}{{dA}}$

For this problem $$D$$ is the region sketched below.

Show Step 2

We’ve done enough double integrals by this point that it should be pretty obvious the best order of integration is to integrate with respect to $$y$$ first.

Here are the limits for the integral with this order.

$\begin{array}{c} - \sqrt 2 \le x \le \sqrt 2 \\ {x^2} - 3 \le y \le 1 - {x^2}\end{array}$

The $$x$$ limits can easily be found by setting the two equations equal and solving for $$x$$.

Show Step 3

The integral for the area is then,

$A = \iint\limits_{D}{{dA}} = \int_{{ - \sqrt 2 }}^{{\sqrt 2 }}{{\int_{{{x^2} - 3}}^{{1 - {x^2}}}{{\,dy}}\,dx}}$ Show Step 4

Now all we need to do is evaluate the integral. Here is the $$y$$ integration.

$A = \iint\limits_{D}{{dA}} = \int_{{ - \sqrt 2 }}^{{\sqrt 2 }}{{\left. y \right|_{{x^3} - 3}^{1 - {x^2}}\,dx}} = \int_{{ - \sqrt 2 }}^{{\sqrt 2 }}{{4 - 2{x^2}\,dx}}$ Show Step 5

Finally, the $$x$$ integration and hence the area of $$D$$ is,

$A = \left. {\left( {4x - \frac{2}{3}{x^3}} \right)} \right|_{ - \sqrt 2 }^{\sqrt 2 } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{16\sqrt 2 }}{3}}}$