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Evaluate the given integral by changing to polar coordinates. double integral_D x^2 y dA, where D...

Question:

Evaluate the given integral by changing to polar coordinates.

{eq}\displaystyle \iint_D x^2 y\ dA {/eq}, where {eq}D {/eq} is the top half of the disk with center the origin and radius {eq}5 {/eq}.

Evaluating Double Integral:

The double integral process is defined as integrating the inner portion first and then, integrating the outer portion.

In this case, we are going to convert the double integral into polar coordinates. The formula is {eq}x =r \cos \theta , \ y=r \sin \theta, \ dA=rdrd\theta {/eq}.

Answer and Explanation:

Given, D is the top half of the disk with center at the origin and radius {eq}5 {/eq}. So, we consider half disk as semicircle.

{eq}\begin{align*} \\ \text{Converting the integral into polar coordinates}: \\ \\ \text{Limits}: & \Rightarrow 0 \leq r \leq 5, \ \ 0 \leq \theta \leq \pi \\ \\ x &=r \cos \theta , \ y=r \sin \theta, \ dA=rdrd\theta \\ \\ \text{I} &= \iint_D x^2 y \ dA \\ \\ &= \int_{0}^{\pi}\int_{0}^{5} \left( (r \cos \theta)^2 r \sin \theta \right) \ rdrd\theta \\ \\ &= \int_{0}^{\pi}\int_{0}^{5} \left( r^4\cos ^2\left(\theta \right)\sin \left(\theta \right) \right) \ drd\theta \\ \\ &=\int_{0}^{\pi} \left( \cos ^2\left(\theta \right)\sin \left(\theta \right) \left[ \frac{r^5}{5} \right]_{0}^{5} \right) \ d\theta \\ \\ &=\int_{0}^{\pi} \left( \cos ^2\left(\theta \right)\sin \left(\theta \right) \left[ \frac{(5)^5}{5}-\frac{(0)^5}{5} \right] \right) \ d\theta \\ \\ &=\int_{0}^{\pi} \left(625 \cos ^2\left(\theta \right)\sin \left(\theta \right) \right) \ d\theta \\ \\ &=625 \int_{0}^{\pi} \left( \cos ^2\left(\theta \right)\sin \left(\theta \right) \right) \ d\theta \\ \\ \mathrm{Using\:u-substitution:}\:u &=\cos \left(\theta \right), \: du=\sin \left(\theta \right) \ d\theta \\ \\ &=625\int_{0}^{\pi} \:-u^2du \\ \\ &=625\left[ -\frac{u^{2+1}}{2+1} \right]_{0}^{\pi} \\ \\ &=625\left[ -\frac{u^{3}}{3} \right]_{0}^{\pi} \\ \\ \mathrm{Substituting,\:}\:u &=\cos \left(\theta \right) \\ \\ &=625\left[ -\frac{(\cos \left(\theta \right))^{3}}{3} \right]_{0}^{\pi} \\ \\ &=625\left[ \left( -\frac{(\cos \left( \pi \right))^{3}}{3} \right)-\left( -\frac{(\cos \left(0 \right))^{3}}{3} \right) \right] \\ \\ &=625\left( \frac{2}{3} \right) \\ \\ &=\frac{1250}{3} \\ \\ \text{I} &=416.67 \end{align*} {/eq}


The answer is {eq}\ \text{I} =416.67 {/eq}.


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Double Integration: Method, Formulas & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 15
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