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Calculate using polar coordinates: $$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$ where D is the region inside the circle with radius 1.

Working:

D: $ \ x^2+y^2=1 \\ $ so

$ 0 \leq r \leq 1 \ \ , $ $0 \leq \theta \leq \pi \ $

and $x=rcos\theta \ ,$ $y=rsin\theta $ so $$\int_{0}^{1} \int_{0}^{\pi} r^2(cos^2\theta + sin^2\theta)^\frac{1}5 \ d\theta \ dr $$ But here is where I get stuck! Isn't $ (cos^2\theta + sin^2\theta) = 1$? So its just
$$\int_{0}^{1} \int_{0}^{\pi} r^2 \ d\theta \ dr $$ I am really confused as what to do! Any help is appreciated

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  • $\begingroup$ where did the $r$ go in the second integral? Also $dx \, dy \neq dr \, d\theta$. $\endgroup$ – Gregory Nov 19 '15 at 17:47
  • $\begingroup$ $dx dy=r dr d\theta$ $\endgroup$ – E.H.E Nov 19 '15 at 17:49
  • $\begingroup$ using your change of variables $x^2+y^2=r^2$, and $\theta$ should go to $2 \pi$. $\endgroup$ – rVitale Nov 19 '15 at 17:49
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According to your calculations it should go on as: $$\iint_{D}^{} (x^2+y^2)^\frac{1}5 \ dx \ dy $$ $$=\int_{0}^{1} \int_{0}^{2\pi} (r^2\cos^2\theta + r^2\sin^2\theta)^\frac{1}{5} \cdot r d\theta \ dr $$ $$=\int_{0}^{1} \int_{0}^{2\pi} r^\frac{2}{5} \cdot r d\theta \ dr$$ $$=\int_{0}^{1} r^\frac{7}{5} dr\int_{0}^{2\pi} d\theta $$ $$=\frac{5}{12} \cdot 2\pi$$ $$=\frac{5\pi}{6}$$

NOTE: The limits for $\theta$ must be $0$ to $2\pi$ and not $0$ to $\pi$.

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  • $\begingroup$ This is so helpful thank you! $\endgroup$ – sarahbenoist Nov 19 '15 at 18:00
  • $\begingroup$ @sarahbenoist You're welcome. Is my answer correct? Can you confirm? $\endgroup$ – SchrodingersCat Nov 19 '15 at 18:01
  • $\begingroup$ yes, it is correct! $\endgroup$ – sarahbenoist Nov 19 '15 at 18:04
  • $\begingroup$ @sarahbenoist If you like my answer, you can click the tick mark beside my answer and accept it. $\endgroup$ – SchrodingersCat Nov 19 '15 at 18:06
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When converting to polar coordinates we make the substitution:

$$x = r\cos(\theta)$$ $$y = r\sin(\theta)$$ $$dA = rdrd\theta$$

With this the integral becomes:

$$\iint_D (x^2 + y^2)^\frac{1}{5} dA = \int_0^{2\pi} \int_0^1 (r^2)^\frac{1}{5} rdrd\theta$$

$$=\int_0^{2\pi} \int_0^1 r^\frac{7}{5} drd\theta$$

$$=\int_0^{2\pi} d\theta \int_0^1 r^\frac{7}{5} dr$$

$$=2\pi \cdot \Big[ \frac{5}{12}r^\frac{12}{5} \Big|_0^1 \Big]$$

$$= \frac{5\pi}{6}$$

where splitting up the double integral into two separate integrals is allowed since we are integrating over a "rectangular region".

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  • $\begingroup$ Check your answer. My limits are 100% correct. $\endgroup$ – SchrodingersCat Nov 19 '15 at 18:00
  • $\begingroup$ You're right, since the region is the whole unit disk the angle has to go all the way around. I made the mistake of following the limits that were given above. $\endgroup$ – Nathan Marianovsky Nov 19 '15 at 18:02

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