Question concerning definite integrals. Lets say we have some integral in cartesian coordinates (like the integral of

$$ \int^{3/4}_0 \sqrt{1+y^2} \space dy$$

I completely understand how to convert the function itself into polar coordinates as well as how to find the indefinite integral of a polar equation. What I do not understand is how to reevaluate the limits of integration. Any help would be appreciated.

  • $\begingroup$ What is your interpretation of this integral "in Cartesian coordinates"? Is it the area of the region bounded by $x=0$, $y=0$, $y=\frac34$, and $x = \sqrt{1+y^2}$, or something else? (And I second the request to see your conversion to polar coordinates, which could also give some clues.) $\endgroup$
    – David K
    Mar 11, 2015 at 4:12

3 Answers 3


Normally an integral in Cartesian coordinates is a two dimensional integral over some region of the plane, not a one dimensional interval as in your example. If it is a simple rectangular region $\int_a^b\int_c^d f(x,y)dx\ dy$ it is hard to express in polar coordinates. That is a strong reason to stay in Cartesian coordinates. If the region is simple in polar coordinates, like a circle at the origin $\int_{-R}^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} f(x,y) dx \ dy$ it becomes simple in polar coordinates $\int_0^{2 \pi}\int _0^R f(r,\theta)r\; dr\; d\theta$ where I have abused notation by using $f$ for both versions, irrespective of the arguments. The second is much simpler for circular regions, so is a candidate when the region is simple in polar coordinates.

  • 1
    $\begingroup$ very right said. $\endgroup$
    – Adnan
    Mar 11, 2015 at 4:21
  • $\begingroup$ @Adnan: Congratulations on your very first real comnent! $\endgroup$ Mar 11, 2015 at 7:17

Generally, re-expressing bounds of an integral in polar coordinates is easier with multiple integrals, but this problem's not impossible.

I've made a rough attempt at creating a picture of the situation, and I'll be referencing that throughout this answer.

When you convert your integral into polar coordinates, you're going to end up integrating with respect to $\theta$, so your bounds should be in terms of $\theta$ as well. As the region currently stands, that's impossible; if $\theta$ were to vary from $0$ to $\frac{\pi}{2}$, you'd need a single function in terms of $r$ that would describe both the smooth curve on the right side of the region and the horizontal line bounding the top of the region. That's either impossible or not worth our time. However, you can split the region up into two smaller regions: a triangle (whose area we can easily calculate with geometry) and a region where you can express $\theta$ as a function of $r$. Adding the areas of each of these two smaller regions will give you the answer you're looking for.

Since the line I've drawn from the origin intersects the graph of $x=\sqrt{1+y^2}$ at the point $\left(\frac{5}{4},\frac{3}{4}\right)$, we can calculate the area of the triangle to be $A_1=\frac{1}{2}\left(\frac{5}{4}*\frac{3}{4}\right)=\frac{15}{32}$.

In the less geometrically simple region, $\theta$ ranges from $0$ to the angle that the hypotenuse of the triangle makes with the origin, which for our purposes is $\frac{\pi}{2}-\arctan{\frac{5}{3}}$. Now we'll need to convert $x=\sqrt{1+y^2}$ to polar coordinates, which I know you said you know how to do, but which I'll nonetheless show here for completeness. $$\begin{align}x&=\sqrt{1+y^2}\\ r\cos\theta&=\sqrt{1+r^2\sin^2\theta}\\ r^2\cos^2\theta&=1+r^2\sin^2\theta\\ r^2\left(\cos^2\theta-\sin^2\theta\right)&=1\\ r^2\cos2\theta&=1\\ r&=\sqrt{\sec2\theta}\end{align}$$

The general form of a single polar integral is $\frac{1}{2}\int_{\theta_1}^{\theta_2}\left[r\left(\theta\right)\right]^2d\theta$, so, plugging in the specifics of this problem, we get the following: $$A_2=\frac{1}{2}\int_0^{\frac{\pi}{2}-\arctan{\frac{5}{3}}}\sec2\theta\,d\theta$$ Making the substitution $\phi=2\theta$, we have $$\begin{align}A_2&=\frac{1}{4}\int_0^{\pi-2\arctan{\frac{5}{3}}}\sec\phi\,d\phi\\ &=\frac{1}{4}\Big[\ln\big\lvert \sec\phi+\tan\phi\,\big\rvert\Big]_0^{\pi-2\arctan{\frac{5}{3}}}\\ &=\frac{1}{4}\ln\Big\lvert\sec\left(\pi-2\arctan{\frac{5}{3}}\right)+\tan\left(\pi-2\arctan{\frac{5}{3}}\right)\Big\rvert-\frac{1}{4}\ln\big\lvert \sec0+\tan0\,\big\rvert\\ &=\frac{1}{4}\ln\Big\lvert-\sec\left(2\arctan{\frac{5}{3}}\right)-\tan\left(2\arctan{\frac{5}{3}}\right)\Big\rvert-\frac{1}{4}\ln\left( 1+0\right)\\ &=\frac{1}{4}\ln\Big\lvert\sec\left(2\arctan{\frac{5}{3}}\right)+\tan\left(2\arctan{\frac{5}{3}}\right)\Big\rvert\end{align}$$ Sparing you a hideous application of some double-angle formulas, that simplifies to $$A_2=\frac{1}{2}\ln2$$ Therefore, the total area is $$\int^{3/4}_0\sqrt{1+y^2}\,dy=A_1+A_2=\frac{15}{32}+\frac{1}{2}\ln2$$


@Louie I would say, first of all you have to make it double integral in order to use substitution using polar coordinate. For your case you can assume

$$ I = \int^{3/4}_0 \sqrt{1+y^2} \space dy$$ or it can be $$ I = \int^{3/4}_0 \sqrt{1+x^2} \space dy$$

both are same as only x variable is replaced by y. Now, multiply both of the above equations, you would get $$ I^2 = \int^{3/4}_0 \int^{3/4}_0 \sqrt{1+x^2} \sqrt{1+y^2} \space dxdy$$ or $$ I^2 = \int^{3/4}_0 \int^{3/4}_0 \sqrt{1+x^2+y^2+x^2y^2} \space dxdy$$

Now use the polar coordinate substitution as


and the limit of one integral will be for $\theta$ from $0$ to $2\pi$ and for the other integral it will be from $0$ to $r$.

After evaluation, take sqaure root of $I^2$ to end up with your integral solution which is for

$$ I = \int^{3/4}_0 \sqrt{1+y^2} \space dy$$

I hope it would help you.

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    $\begingroup$ No, because the region when doing $I^2$ in Cartesian coordinates is a square and the one you imply in polar coordinates is a circle. $\endgroup$ Mar 11, 2015 at 4:45
  • $\begingroup$ @Adnan Have you actually tried doing the integral you're suggesting? Give it a shot; see what happens. $\endgroup$ Dec 10, 2017 at 20:31

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