Question

Evaluate

$\displaystyle \int\limits_{0}^{1} dx \int\limits_{0}^{\sqrt{1-x^{2}}} \frac{dy}{(1+e^{y})\sqrt{1-x^{2}-y^{2}}}$

by changing the order of integration.​​

Answer 1

We will be using the Fubini's Theorem.

Fubini's Theorem

If f is a continuous function on a Rectangle R: [a,b] $\times$ [c,d], then

$\boxed{\sf\displaystyle \iint_R f(x,y)\, dx\, dy = \int\limits_c^d \left( \int\limits_a^b f(x,y) \, dx \right) \,dy = \int\limits_a^b \left( \int\limits_c^d f(x,y) \, dy \right) \,dx }$

Fubini's Theorem helps us change the order of integrals without affecting the value of integral.

Here, the limits of x are from 0 to 1, while the limits of y go from 0 to $\sqrt{1-x^2}$.

It is clear that integrating this is going to be difficult. However, if we are able to change the order of integrals, the integration should become easier. Let's proceed to do just that!

It is fairly intuitive to see that this integration is being done over a domain which is a circle. And a circular domain is a continuous one. So, we can and will use the Fubini's Theorem.

To change the order, we must bring the limits of x in terms of y.

The limits of y are from 0 to $\bf\sqrt{1-x^2}$.

Consider:

$y=\sqrt{1-x^2} \\ \\ \\ \implies y^2=1-x^2 \\ \\ \\ \implies x^2+y^2 = 1 \quad \textsf{which is a circle} \\ \\ \\ \implies x = \sqrt{1-y^2}$

See Graphs 1 and 2.

Graph 1 shows the original integral of value of y ranging from 0 to $\sqrt{1-x^2}$. So, it is a semicircle in the Positive y direction.

This inner integral can be visualized as the little green strip moving horizontally along the x axis. This strip represents the part of domain.

Also, the limits of x are from 0 to 1. This reduces the domain of the integral.

Now, we surely know that the domain of the integral is a unit circle in the first quadrant.

Now, see Graph 2. It shows the order of integrals changed.

We see that by changing limits of x as x from 0 to $\sqrt{1-y^2}$, the limits of y also change. The new limits are y = 0 to y = 1.

This new inner integral in form of x can be visualized as a narrow strip running vertically, with y varying from 0 to 1.

Hence, we can now calculate the integral:

$\sf\displaystyle \int\limits_{0}^{1} dx \int\limits_{0}^{\sqrt{1-x^{2}}} \frac{dy}{(1+e^{y})\sqrt{1-x^{2}-y^{2}}}\\\\\\ =\int\limits_{0}^{1} dy \int\limits_{0}^{\sqrt{1-y^{2}}} \frac{dx}{(1+e^{y})\sqrt{1-x^{2}-y^{2}}}\\\\\\ =\int\limits_{0}^{1} \frac{dy}{1+e^y} \int\limits_{0}^{\sqrt{1-y^{2}}} \frac{dx}{\sqrt{\left(\sqrt{1-y^2}\right)^2-x^{2}}}$

$\boxed{\begin{minipage}{15 em} \textsf{Use the identity:}\\ \\ \displaystyle\sf \int \frac{dx}{\sqrt{x^2-a^2}} = \sin^{-1} \left(\frac{x}{a} \right) + c\\ \\ \\ \textsf{Here we have a de\ \!\!finite integral,}\\\textsf{so +c won't occur} \end{minipage}}$

$\displaystyle\sf = \int\limits_{0}^{1} \frac{dy}{1+e^y}\left[\sin^{-1} \left(\frac{x}{\sqrt{1-y^2}}\right) \right]_{0}^{\sqrt{1-y^2}}\\\\\\ = \int\limits_{0}^{1} \frac{dy}{1+e^y}\left[\sin^{-1} \left(\frac{\sqrt{1-y^2}}{\sqrt{1-y^2}}\right)-\sin^{-1} \left(\frac{0}{\sqrt{1-y^2}}\right)\right]\\\\\\ = \int\limits_{0}^{1} \frac{dy}{1+e^y}\left[\sin^{-1} (1)-\sin^{-1} (0) \right] \\\\\\ = \int\limits_{0}^{1} \frac{dy}{1+e^y}\left(\frac{\pi}{2}-0\right)$

$\sf\displaystyle = \frac{\pi}{2} \int\limits_{0}^{1} \frac{dy}{1+e^y}\\\\\\ \boxed{\begin{minipage}{10em} \sf\textsf{Put }1+e^y=t\\\\\implies e^y \, dy = dt \\ \\ \implies dy = \dfrac{dt}{e^y} = \dfrac{dt}{t-1} \end{minipage}}\:\boxed{\begin{minipage}{15em} \sf \textsf{Regarding the Limits}\\ \\ At\ y=0, t=e^0+1=1+1=\bold{2} \\ \\ \\ At\ y=1, t=e^1+1=\bold{e+1} \end{minipage}}\\\\\\ = \frac{\pi}{2} \int\limits_2^{e+1} \frac{dt}{t(t-1)}\\\\\\ \textsf{Let's split this as Partial Fractions}$

$\sf\displaystyle = \frac{\pi}{2} \int\limits_2^{e+1} \frac{dt}{t(t-1)}\\\\\\ = \frac{\pi}{2} \int\limits_2^{e+1}\left( \frac{t-(t-1)}{t(t-1)}\right) dy \\\\\\ = \frac{\pi}{2}\int\limits_2^{e+1}\left( \frac{1}{t-1}-\frac{1}{t}\right)dy\\\\\\ = \frac{\pi}{2}\left(\int\limits_2^{e+1}\frac{dy}{t-1}-\int\limits_2^{e+1}\frac{dy}{t}\right)\\\\\\ = \frac{\pi}{2} \left[ \ln (t-1) - ln(t) \right]_2^{e+1} \\\\\\ = \frac{\pi}{2}\left[\ln\left(\frac{t-1}{t}\right)\right]_2^{e+1}}$

$\sf\displaystyle =\frac{\pi}{2}\left[\ln\left(\frac{e+1-1}{e+1}\right)-\ln\left(\frac{2-1}{2}\right)\right]\\\\\\ = \frac{\pi}{2}\left[\ln\left(\frac{e}{e+1}\right)-\ln\left(\frac{1}{2}\right)\right]\\\\\\ = \frac{\pi}{2}\ln\left(\frac{2e}{e+1}\right)$

Thus,

$\Large\boxed{\sf\displaystyle\int\limits_{0}^{1} dx \int\limits_{0}^{\sqrt{1-x^{2}}} \frac{dy}{(1+e^{y})\sqrt{1-x^{2}-y^{2}}} = \frac{\pi}{2}\ln\left(\frac{2e}{e+1}\right)}$

Hence, we see that changing the order of integrals by Fubini's Theorem can greatly simplify calculations, and helps to find the value of the integrals as well.

Answer 2

Sometimes you need to change the order of integration to get a tractable integral. For example, if you tried to evaluate

∫10∫1xey2dydx

directly, you would run into trouble. There is no antiderivative of ey2, so you get stuck trying to compute the integral with respect to y. But, if we change the order of integration, then we can integrate with respect to x first, which is doable. And, it turns out that the integral with respect to y also becomes possible after we finish integrating with respect to x.

According to the limits of integration of the given integral, the region of integration is

0≤x≤1x≤y≤1,

which is shown in the following picture.

Change order of integration example triangular region

Since we can also describe the region by

0≤y≤10≤x≤y,

the integral with the order changed is

∫10∫1xey2dydx=∫10∫y0ey2dxdy

With this new dxdy order, we integrate first with respect to x

∫10∫y0ey2dxdy=∫10xey2∣∣x=yx=0dy=∫10yey2dy.

Since the integration with respect to x gave us an extra factor of y, we can compute the integral with respect to y by using a u-substitution, u=y2, so du=2ydy. With this substitution, u rannges from 0 to 1, and we calculate the integral as

∫10∫y0ey2dxdy=∫10yey2dy=∫1012eudu=12eu∣∣∣10=12(e−1).