# Area under the curve

## Main Question or Discussion Point

how do you find the integral of the equation sqrt(36-.22x^2) between x=0 and x=9

## Answers and Replies

arildno
Homework Helper
Gold Member
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Well, we have:
$$\int_{0}^{9}\sqrt{36-0.22x^{2}}dx=6\int_{0}^{9}\sqrt{1-(\frac{\sqrt{0.22}x}{6})^{2}}dx$$

Set:
$$u=\frac{\sqrt{0.22}x}{6}}\to{dx}=\frac{6}{\sqrt{0.22}}$$
And our integral may be rewritten as:
$$\frac{36}{\sqrt{0.22}}\int_{0}^{\frac{3\sqrt{0.22}}{2}}\sqrt{1-u^{2}}du$$

Do you have any ideas how to proceed from here?