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I know that the area of a circle, $x^2+y^2=a^2$, in cylindrical coordinates is $$ \int\limits_{0}^{2\pi} \int\limits_{0}^{a} r \, dr \, d\theta = \pi a^2 $$

But how can find the same result with a double integral and only cartesian coordinates?

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Think about how the cartesian variables $x$ and $y$ are bounded. If we have the equation $$ x^2+y^2=r^2\Rightarrow x=\pm\sqrt{r^2-y^2}\;\text{or}\;y=\pm\sqrt{r^2-x^2} $$ And $|x|,|y|<r$. Note that this last condition also insures that the above square roots are real. Then this gives you bounds for your double integral, choosing to integrate $x$ first, $$ \int_{-r}^r\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\mathrm dy= \int_{-r}^r2\sqrt{r^2-y^2}\mathrm dy $$ Which you can integrate using the substitution $y=r\sin(t)\Rightarrow \mathrm dy=r\cos(t)\mathrm dt$ $$ 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2\cos^2(t)\mathrm dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}r^2(1+\cos(2t))\mathrm dt\\ =r^2(t+\frac{1}{2}\sin(2t))\vert_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=r^2\pi $$ Note that I used the identity $\cos^2(t)=1/2(1+\sin(2t))$ to evaluate.

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  • $\begingroup$ Great, thanks! Just for curiosity, is it even possible with this substitution? $$t= r^2-y^2$$ $$\frac{dt}{dy}=-2y$$ $$dt=-2ydy$$ And $y^2=r^2-t, y=\pm \sqrt{r^2-t}$ so $$dt=-2\sqrt{r^2-t}dy$$ $$\frac{-dt}{\sqrt{r^2-t}}=2dy$$ Stuck! $\endgroup$ – JDoeDoe Jul 18 '16 at 18:03
  • $\begingroup$ at first glance it looks like it may work, try pulling out $r^2$ and then use substitution $u=t/r^2$ to see arcsin $\endgroup$ – qbert Jul 18 '16 at 18:09
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$$I=\int_{-r}^r \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} dy \, dx\\ \\ I=\int_{-r}^r 2\sqrt{r^2-x^2}\, dx\\ \\ $$ Set $x=r \sin t$, so $dx = r \cos t\,dt\,$, we have $$ I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2r^2 \cos^2 t \,dt\\ \\ r^2(t+\sin t \cos t)\Big{|}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}= \color{red}{\pi r^2}$$

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Just as an alternative solution to the qbert answer.

Note that $|x|$, $|y|$ are not strictly less than $r$, but instead $|x| \leq r$, $|y| \leq r$. This can still insure that $r^2 \geq y^2$ and so $\sqrt{r^2-y^2}$ is real.

After the first integral

$$ \int_{-r}^r\int_{-\sqrt{r^2-y^2}}^{\sqrt{r^2-y^2}}\mathrm dx\mathrm dy= \int_{-r}^r2\sqrt{r^2-y^2}\mathrm dy $$

Instead of the change of variable, you can remember the known integral:

$$\int \sqrt{r^2 - y^2} = \frac{1}{2} \left( r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \right) + C$$

whose condition $|y| \leq |r|$ has just been mentioned and it is satisfied. The result follows almost immediately:

$$\int_{-r}^r 2 \sqrt{r^2 - y^2} \mathrm{d} y = \left[ r^2 \arcsin \frac{y}{r} + y \sqrt{r^2 - y^2} \ \right]_{-r}^r = r^2 \arcsin (1) - r^2 \arcsin (-1) = \\ = r^2 \frac{\pi}{2} + r^2 \frac{\pi}{2} = \pi r^2$$

So, using the cartesian coordinates, the only important observation is simply to consider the right extreme values for each variable.

Using the circumference equation $x^2 + y^2 = r^2$, you can choose $x$ as a function of $y$, obtaining $x = \pm \sqrt{r^2 - y^2}$, which will be the extreme values for $x$. Then, you let $y$ sweep from $-r$ to $r$, which are the extreme values for $y$. Of course, you can alternatively do vice-versa, with $y$ as function of $x$.

Unlike the cylindrical coordinates, there is no angular variation here, but only an horizontal variation between $- \sqrt{r^2 - y^2}$ and $\sqrt{r^2 - y^2}$ as regards $x$, and a vertical variation between $-r$ and $r$ as regards $y$.

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I know this is not in the context of the question, but it seems as though you asked this question out of curiosity. So actually, using Green's Theorem you can get the result with the single integral,

$$\frac{1}{2} \oint_C x \ dy - y \ dx$$

where $C$ is the underlying curve for the parametrization $\gamma(t) = (R \cos t , R \sin t)$ and $0 \leq t\leq 2 \pi$.

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Do you mean something like this?

$$\int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{+\sqrt{a^2-x^2}} 1 \, dy \, dx$$

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  • $\begingroup$ Yes, how do you evaluate that without cylindrical coordinates? $\endgroup$ – JDoeDoe Jul 18 '16 at 17:53
  • $\begingroup$ I don't understand your question. These are Cartesian coordinates. $\endgroup$ – Martin Kochanski Jul 19 '16 at 8:28

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