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Ernstfalld

Answered question

2021-10-12

Evaluate the iterated integral by converting to polar coordinates.

\int_{0}^{2} \int_{0}^{\sqrt{(2 x - x^{2})}} x y d y d x

Leonard Stokes

Skilled2021-10-13Added 98 answers

Step1
Integrating the given equation after converting to polar coordinates:

\int_{0}^{2} \int_{0}^{\sqrt{(2 x - x^{2})}} x y d y d x

Let is assume x = r \cos (θ) and y = r \sin (θ) after putting in above expressiona nd corresponding to that changing the limiting value:

Expression will change into

I = \int_{0}^{{\frac{π {2}}{\int}}_{0}^{2 \cos θ} r^{3} \cos (θ) \sin (θ) d r d θ}

= \int_{0}^{\frac{π}{2}} \cos (θ) \sin θ \cdot [\int_{0}^{2 \cos θ} r^{3} d r] d θ

= \int_{0}^{\frac{π}{2}} \cos^{5} (θ) \sin (θ) d θ

Now assuming = \cos (θ) so d u = - \sin (θ)

Putting this to solve further

I = - \int_{0}^{\frac{π}{2}} u^{5} d u

= {[\frac{- 4 \cos^{6} (θ)}{6}]}_{0}^{\frac{π}{2}}

= \frac{2}{3}

Result:

= \frac{2}{3}

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