area under the curve

[ngigs]

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

[arildno]

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?