Answered: a (0 32. x²y dx dy -Va?-y?

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Evaluate the iterated integral by converting to polar coordinates.

Expert Solution

Step 1

Given Integral, $I = \int_{0}^{a} \int_{- \sqrt{a^{2} - y^{2}}}^{0} x^{2} y d x d y$

Use Polar coordinates,

$x = r \cos θ, y = r \sin θ, d x d y = r d r d θ l i m i t s : x = 0 \Rightarrow r \cos θ = 0 \Rightarrow θ = \frac{π}{2} x = - \sqrt{a^{2} - y^{2}} \Rightarrow x^{2} + y^{2} = a^{2} \Rightarrow r = a x = - \sqrt{a^{2} - y^{2}} \Rightarrow x = - a \sqrt{1 - \sin^{2} θ} \Rightarrow x = - a \cos θ \Rightarrow a = - a \cos θ \Rightarrow θ = π$

Step 2

Thus,

$\begin{array}{rcl} \int_{0}^{a} \int_{- \sqrt{a^{2} - y^{2}}}^{0} x^{2} y d x d y & = & \int_{0}^{a} \int_{\frac{π}{2}}^{π} {(r \cos θ)}^{2} (r \sin θ) r d r d θ \\ = & \int_{0}^{a} \int_{\frac{π}{2}}^{π} r^{4} \cos^{2} θ \sin θ d r d θ \\ = & \int_{0}^{a} r^{4} d r \int_{\frac{π}{2}}^{π} \cos^{2} θ \sin θ d θ \\ = & \int_{0}^{a} r^{4} d r [- \int_{\frac{π}{2}}^{π} \cos^{2} θ (- \sin θ) d θ] \\ = & \int_{0}^{a} r^{4} d r {[\frac{- \cos^{3} θ}{3}]}_{\frac{π}{2}}^{π} \\ = & {[\frac{r^{5}}{5}]}_{0}^{a} [\frac{- \cos^{3} (π)}{3} + \frac{- \cos^{3} (\frac{π}{2})}{3}] \\ = & \frac{a^{5}}{5} [\frac{1}{3} + 0] \\ \int_{0}^{a} \int_{- \sqrt{a^{2} - y^{2}}}^{0} x^{2} y d x d y & = & \frac{a^{5}}{15} \end{array}$

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