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I'm working on this question for my Calculus III Homework:

Evaluate the given integral by changing to polar coordinates.

$$\iint_{R} (5x-y)\,dA$$

where R is the region in the first quadrant enclosed by the circle $x^2 + y^2 = 16$ and the lines $x = 0$ and $y = x$.

I mapped out the coordinates and got $\displaystyle\iint_R (5r\cos\theta-r\sin\theta)\,r \,dr\, d\theta$, where $0 \le r \le 4$ and $0 \le \theta \le \pi/4 $. Working it out it came out to $64 \sqrt{2} \, 64/3$, which was incorrect. If anyone could point out where I went wrong (most likely with defining coordinates), I would appreciate it very much.

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    $\begingroup$ Your question is a bit ill formatted, but at first sight it seems you may have gotten the volume element wrong. Note that $dA = rdr d\theta$ $\endgroup$ – Thomas Oct 11 '15 at 19:47
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    $\begingroup$ It is $\pi/4\le\theta\le\pi/2$ $\endgroup$ – Aretino Oct 11 '15 at 19:47
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    $\begingroup$ Can you explain how you got this theta value? I don't understand. $\endgroup$ – TomShoe Oct 11 '15 at 19:59
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    $\begingroup$ The line $x=0$ corresponds to the $y$-axis, and in the first quadrant that means $\theta=\pi/2$. $\endgroup$ – Aretino Oct 11 '15 at 20:01
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    $\begingroup$ You can use the standard code \iint instead of \int\!\!\!\!\int. I changed it. Also, for your first one, I used an actual display and deleted the \displaystyle. $\endgroup$ – Michael Hardy Oct 11 '15 at 20:19
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$$\int_{\pi/4}^{\pi/2}\int_{0}^{4}(5r\cos\theta-r\sin \theta)r\,dr\,d\theta$$ $$\int_{\pi/4}^{\pi/2}\int_{0}^{4}(5r^2\cos\theta-r^2\sin \theta)\,dr\,d\theta$$ $$\int_{\pi/4}^{\pi/2}(5\cdot64/3\cos\theta-64/3\sin \theta)\,d\theta$$ $$(5\cdot64/3(1-\frac{1}{\sqrt{2}})+64/3(0-\frac{1}{\sqrt{2}})=\frac{320-192\sqrt{2}}{3}$$ enter image description here

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$$ \int \left( \int (r\cos\theta) r\,dr\right)\,d\theta = \int\left( (\cos\theta) \int r^2\,dr \right) \, d\theta $$ The above can be done because $\cos\theta$ does not change as $r$ changes; thus it is a "constant" as far as the inside integral is concerned.

Next, observe that the inside integral, with respect to $r$, does not change as $\theta$ changes, since no $\theta$ appears in it. It is therefore a "constant" as far as the outside integral is concerned, and thus can be pulled out, getting $$ \int \cos\theta\,d\theta \cdot \int r^2\,dr. $$

As for the bounds on $\theta$: draw the picture. The line $y=x$ is where $\theta=\pi/4$, and the line $x=0$ is where $\theta=\pi/2$.

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