# Converting cartesian to polar double integral

Convert the following integral to polar coordinates. You do not need to evaluate. $$\int_{-3}^3 \int_{x}^{\sqrt{9-x^2}} x^2y dy dx$$

My work :

I plotted the limits and I don't understand the bounded region due to $$y=x$$, but still I got like this which is wrong I know I solved the integral it should be $$\frac{-81}{5}$$ but the integral in the polar coordinates I obtained is incorrect $$\int\limits_{\pi/4}^{\pi}\int\limits_{0}^{3}r^4\cos^2 \theta \sin \theta dr d\theta+ \int\limits_{\pi}^{5\pi/4}\int\limits_{-3/\cos \theta}^{-3\sqrt2}r^4\cos^2 \theta \sin \theta dr d\theta$$

Can anyone help me recorrecting it the answer below is not complete, and it is definitely not the double of the answer ???

• From a sketch it appears that in polar coordinates you would be required to seperate the integral into $3$ sub integrals which add to the original integral. – Peter Foreman May 29 at 7:22
• Have you drawn the region? Seen what it actually looks like in the plane? That is usually very helpful when converting integral bounds. – Arthur May 29 at 7:23
• I have tried to correct some typos and also copy the text of the problem - so that it isn't included as a picture. Here is link to the current revision. I am not really sure why the previous edit (of the same nature) was removed. – Martin Sleziak May 29 at 9:23
• @learningstudent It's not just about grammar. (But even if it was, I do not really see why the edit was reverted.) Another thing, which you probably know by know, is that for several reasons including problem just as a picture is unsuitable. For example: What should I do when I see a “pic-question”? and other posts linked there. – Martin Sleziak May 29 at 9:33

The integration area is shaded and their polar boundaries are marked in the graph. Moreover, in polar coordinates $$x^2y dy dx= r^4\cos^2\theta\sin\theta drd\theta=f(r,\theta)drd\theta$$ and the integral is given as

$$\int_{\frac\pi4}^0 \int_3^{\frac3{\cos\theta}} f(r,\theta) drd\theta +\int_{\frac\pi4}^{\pi} \int_0^{3} f(r,\theta)drd\theta +\int_{\pi}^{\frac{5\pi}4} \int_0^{-\frac3{\cos\theta}} f(r,\theta)drd\theta = - \frac{81}5$$

• A negative answer? You can tell the answer is positive by oddness in the region where $3\pi/4\le \theta\le 5\pi/4.$ – zhw. Jun 5 at 22:39
• let me know why it is $-3/cos\theta$ ? It should be plus !! – learningstudent Jun 6 at 4:08
• @learningstudent - because $\cos \theta < 0$ within the integration limits $\theta \in (\pi, 5\pi/4)$ – Quanto Jun 6 at 4:14
• ohhhhh so that is why u do pi/4 to 0 instead of 0 to pi/4 – learningstudent Jun 6 at 4:21
• @learningstudent - that’s right – Quanto Jun 6 at 4:36

Remark that $$x\ge \sqrt{9-x^2}$$ when $$\frac 3{\sqrt 2}\le x\le 3$$.

By a well-known theorem (see, for instance, [Fich, 596]), the given double integral is equal to $$\int_{S} x^2y dx dy-\int_{S’} x^2y dx dy$$, where $$S=\left\{(x,y)\in\Bbb R^2: -3\le x\le 3,\, x\le y\le \sqrt{9-x^2}\right\}$$ and $$S’=\left\{(x,y)\in\Bbb R^2: \frac 3{\sqrt 2}\le x\le 3,\, \sqrt{9-x^2}\le y\le x\right\}.$$

The correspondence between Cartesian coordinates $$(x,y)\in\Bbb R^2$$ and polar coordinates $$(r,\varphi)\in\Bbb R^+\times [0,2\pi)$$ is $$x=r\cos\varphi$$, $$y=r\sin\varphi$$, $$S=\{(r,\varphi): 0\le r\le 3,\, \pi/4\le \varphi\le \pi\}\cup$$ $$\{(r,\varphi): 0\le r\le - 3/\cos\varphi,\, \pi\le \varphi\le 5\pi/4\},$$ $$S’=\{(r,\varphi): 3\le r\le 3/\cos\varphi,\, 0\le \varphi\le \pi/4\},$$

and $$dS=r dr d\varphi$$. Thus

$$\int_S x^2y dx dy=\int_{\pi/4}^{\pi}\int_0^3 r^4\cos^2\varphi\sin\varphi dr d\varphi+$$ $$\int_{\pi}^{5\pi/4}\int_0^{-3/\cos\varphi} r^4\cos^2\varphi\sin\varphi dr d\varphi- \int_{0}^{\pi/4}\int_0^{3/\cos\varphi} r^4\cos^2\varphi\sin\varphi dr d\varphi.$$

References

[Fich] Grigoriy Fichtenholz, Differential and Integral Calculus, v. III, 4-th edition, Moscow: Nauka, 1966, (in Russian).

• wrong answer -1 , try to improve , your answer = $+\frac{81}{5\sqrt2}$ – learningstudent Jun 4 at 15:31
• @learningstudent Thanks. Fixed. – Alex Ravsky Jun 4 at 16:23
• So you're looking at the points inside this circle with angles between $\frac\pi4$ and $\frac\pi2$. How do you get that? It should be the angles between $\frac\pi4$ and $\frac {5\pi}{4}$. – zhw. Jun 5 at 22:33
• @zhw. My first impression about the integration domain was the same as in your answer, but it turned out to be wrong, and, as pointed learningstudent above, provided a wrong value of the integral. A correct situation is shown at the picture in Quanto’s answer. – Alex Ravsky Jun 6 at 3:41
• @AlexRavsky Thanks for your comment. Yes, I completely missed the point of this one. – zhw. Jun 6 at 5:06

For $$x\le y \le \sqrt{9-x^2}$$ we get the points inside the circle $$x^2+y^2=3^2$$ (with radius equal to $$3$$) and above the line $$y=x$$. So you're looking at the points inside this circle with angles between $$\frac\pi4$$ and $$\frac\pi2$$. (The angle $$\pi/4$$ corresponds to the points on the line $$y=x$$.)

We need the points which are below the line and outside of the circle. The second integral is with negative sign, since with positive sign we would get the integral from $$\sqrt{9-x^2}$$ to $$x$$.

$$\int_0^3 \int_{x}^{\sqrt{9-x^2}} x^2y dy dx= \int_{\frac\pi4}^\frac\pi2 \int_0^3 r^4 \cos^2\theta\sin\theta dr d\theta- \int_0^{\frac\pi4} \int_3^{3/\cos\theta} r^4 \cos^2\theta\sin\theta dr d\theta.$$

(For the integral on the LHS WolframAlpha returns $$-\frac{81}{10}$$. On the RHS we get this integral and this integral, they together give us $$\frac{81}{10}(\frac1{\sqrt2}-\frac{2+\sqrt2}2)=-\frac{81}{10}$$.)

This returns only the part of integral for $$x\ge0$$. But since the whole integral is symmetric, you can then multiply this by the factor of two to get the integral from the original problem. We can multiply this by the factor of two to get the integral $$\int_{-3}^3 \int_{|x|}^{\sqrt{9-x^2}} x^2y dy dx.$$ We're still missing this part: $$\int_{-3}^0 \int_{x}^{-x} x^2y dy dx.$$ However, this integral will be zero since $$\int_x^{-x} ydy=\left[\frac{y^2}2\right]_x^{-x}=0.$$

• @learningstudent I suppose this part corresponds to the part between the circle and the vertical line: $\int\limits_{0}^{\pi/4}\int\limits_{3/\cos \theta}^{3\sqrt2}r^4\cos^2 \theta \sin \theta dr d\theta$. Shouldn't then the lower bound be $3$ (points on the circle) and the upper bound $3/\cos\theta$ (points on the vertical line $x=3$)? – Martin Sleziak May 29 at 9:54
• @learningstudent With positive sign, I would get an integral which corresponds to $\int_{\sqrt{9-x^2}}^x \dots dy$. If I want $\int_x^{\sqrt{9-x^2}}$, I need the negative sign. (Notice that in this area we have $\sqrt{9-x^2}<x$. – Martin Sleziak May 29 at 10:52
• what about the region from pi/2 to 5pi/4 ?? – learningstudent May 29 at 10:54