Get the answer to "Evaluate the iterated integral by converting to polar..."

Michael Maggard

Answered question

2021-12-15

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

Annie Levasseur

Beginner2021-12-16Added 30 answers

Integrating the given equation after converting to polar coordinates:

I = \int_{0}^{2} \int_{0}^{\sqrt{2 x - x^{2}}} x y d y d x

Let us assume

x = r \cos θ

and

y = r \sin θ

after putting in above expressionand 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

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

Now assuming

u = \cos (θ)

d u = - \sin (θ)

Putting this ti solve further

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

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

= \frac{23}{}

Linda Birchfield

Beginner2021-12-17Added 39 answers

Maybe I can help you with this question later.

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