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Sketch the region of integration and then compute the following double integral: $$\int_0^1 \int_y^{\sqrt{2-y^2}}(x+y)dxdy$$

https://www.desmos.com/calculator/k5uz2mr2ml

Then I characterized the region as: $$R=\{(r,\theta) \,\,\, | \,\,\, 0 \leq r \leq ?? \,\,\,\, \& \frac{\pi}{4} \leq \theta \frac{\pi}{2} \}$$

I thought that the upper bound for r should be $sec(\theta)$ because if $y=x$, we get $$x^2+y^2=2 \implies 2x^2=2 \implies r^2cos^2\theta = 1 \implies r=sec(\theta)$$

But if that is the case, the following integral diverges as $sec(\frac{\pi}{2} )$is not defined.

$$\int_{\pi/4}^{\pi/2} \int_0^{sec(\theta)} r^2(cos(\theta)+sin(\theta)) drd\theta$$

$$=\frac{1}{3}\int_{\pi/4}^{\pi/2} (sec^2(\theta) + \frac{sin(\theta)}{cos^2(\theta)} ) d\theta \implies \frac{1}{3} \left ( tan(\theta) + \frac{1}{2cos^2(\theta)} \right )_{\frac{\pi}{4}}^{\frac{\pi}{2}}= ???? $$

Where did I mess up? How would you approach this problem?

Thank you in advance for your time.

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  • $\begingroup$ What about $r=\sqrt{2}$? $\endgroup$
    – Ángel Mario Gallegos
    Aug 10, 2015 at 2:44
  • $\begingroup$ But r does not vary from $0 \leq r \leq \sqrt2$ for x<1 $\endgroup$
    – CivilSigma
    Aug 10, 2015 at 2:47
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    $\begingroup$ The sketch is not right. The upper bound on $x$ (i.e, the bound on the right) is the circle (and this is why $r$ goes from 0 to $\sqrt 2$), and the lower bound (i.e., the bound on the left) is the line; the region should be the sector of the circle between angles $0$ and $\pi/4$. $\endgroup$
    – coldnumber
    Aug 10, 2015 at 2:49
  • $\begingroup$ @coldnumber OMG! Now it makes sense why the upper bound of r is $\sqrt2$ . Thank you very much! Let's say the region I drew was correct, would my deduced upper bound for r be correct? $\endgroup$
    – CivilSigma
    Aug 10, 2015 at 2:51
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    $\begingroup$ You're welcome :). Then the upper limit would be $1=y=r\sin\theta \implies r=csc\theta$, I think. You'd use the cosine if you were limited by a vertical line. $\endgroup$
    – coldnumber
    Aug 10, 2015 at 2:53

1 Answer 1

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We have

$$\begin{align} \int_0^1\int_y^{\sqrt{2-y^2}}(x+y)dxdy&=\int_0^{\pi/4}\int_0^{\sqrt{2}}(r\cos \theta+r\sin \theta)rdrd\theta\\\\ &=\frac{2^{3/2}}{3} \end{align}$$

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  • $\begingroup$ Why is $r=\sqrt2$ on the upper bound, I don't see it. $\endgroup$
    – CivilSigma
    Aug 10, 2015 at 2:49
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    $\begingroup$ Good question. We have $x=\sqrt{2-y^2}\implies x^2+y^2=(\sqrt{2})^2$, which is the equation of a circle with radius $\sqrt{2}$ $\endgroup$
    – Mark Viola
    Aug 10, 2015 at 2:51
  • $\begingroup$ Thank you, I see it now ( my original sketch was wrong, which threw me off). But let's say the region I drew was correct, would my deduced upper bound for r be correct? $\endgroup$
    – CivilSigma
    Aug 10, 2015 at 2:53
  • $\begingroup$ You're welcome. My pleasure and very pleased to hear that you've got it now! And please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Aug 10, 2015 at 2:55
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    $\begingroup$ A sketch would of really helped, because if it weren't for @coldnumber mentioning that I had the wrong sketch, I would of had a lot of difficulty processing the information due to what I believed. But other than that, your answer is spot on! Thank you Dr. MV . $\endgroup$
    – CivilSigma
    Aug 10, 2015 at 3:01

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