Re: Solve double integral

That has no simple anti-derivative do swap the order of integrals:

First, you need to be careful about the limits of integration. The best way to do that is to draw a "picture". The outer integral is from x= 0 to $\displaystyle x= \sqrt{\pi}$ so draw vertical lines on an xy- coordinate systems at those values of x. The "inner" integral is from y= x to $\displaystyle y= \sqrt{\pi}$ so draw the slant line y= x and the horizontal line $\displaystyle x= \sqrt{\pi}$. You should see that the region you are to integrate over is the upper half of the square with vertices at (0, 0), $\displaystyle (\pi, 0)$, $\displaystyle (\pi, \pi)$, and $\displaystyle (0, \pi)$. Overall, y goes from to $\displaystyle \sqrt{\pi}$ and, for each y, x goes from x= 0 to $\displaystyle x= y$.

Re: Solve double integral

I tried to swap the order of the integrals (like I posted in my update), so

$\displaystyle \int_x^{\sqrt{\pi}} \int_0^{\sqrt{\pi}} \! 2\sin{(y^2)} \, \mathrm{d} x\mathrm{d} y$

But now the outer integral has a variable boundary (x) I can't get rid off.

Moreover, I still need to find the antiderivative of $\displaystyle 2\sin{(y^2)}$.

I need some more help, please!

Re: Solve double integral

Hi,

May be you did not figure out HallsofIvy's idea.

I think his idea is like that, you see that the domain of integration is the triangle whose vertices are the points $\displaystyle (0,0),(\sqrt{\pi},\sqrt{\pi}),(0,\sqrt{\pi}) $.

So the domain can be alternatively determined as follows :

$\displaystyle 0\leq y\leq \sqrt{\pi}, 0\leq x\leq y$.

So

$\displaystyle \int_{0}^{\sqrt{\pi}}\int_{0}^{y}2\sin y^2 dxdy.$ Now it is straightforward.

Re: Solve double integral

Please do not stop after reading just the first sentence!

My last line was:

"Overall, y goes from 0 to $\displaystyle \sqrt{\pi}$, and, for each y, x goes from x= 0 to y."

$\displaystyle 2\int_{y= 0}^{\sqrt{\pi}}\int_{x= 0}^y sin(y^2) dxdy$

Re: Solve double integral

Quote:

Originally Posted by

**HallsofIvy** Did you not see the last line of my previous post?

"Overall, y goes from 0 to $\displaystyle \sqrt{\pi}$, and, for each y, x goes from x= 0 to y."

$\displaystyle 2\int_{y= 0}^\sqrt{\pi}\int_{x= 0}^y sin(y^2) dxdy$

Yes of course, but I think he did not figure out your post because he asked for further explanations.

Re: Solve double integral

Thank you both for your help! I (think I) get it now.

My solution for the double integral is 1. Is this correct?

Re: Solve double integral

Is it not 2? I got 2. do not forget that you already have 2 in the integral.

Re: Solve double integral

I figured the 2 disappears while calculating the last antiderivative because the antiderivative of $\displaystyle 2y\sin(y^2)$ is $\displaystyle -\cos(y^2)$ and $\displaystyle -\cos(\pi) = 1$

**Update:** I forgot the last term, the answer is not $\displaystyle -\cos(\pi) = 1$ but $\displaystyle -\cos(\pi) - (-\cos(0))= 2$

Re: Solve double integral

Quote:

Originally Posted by

**Lotte1990** I figured the 2 disappears while calculating the last antiderivative because the antiderivative of $\displaystyle 2ysin(y^2)$ is $\displaystyle -cos(y^2)$ and $\displaystyle -cos(\pi) = 1$

note also the lower bound of the integral which is zero the answer will be $\displaystyle -(cos(\pi)-\cos0) = 1+1$[/QUOTE]

Re: Solve double integral

You are right! I forgot to substract the lower boundary! I get it now! Thank you so much for your help!

Re: Solve double integral

Quote:

Originally Posted by

**Kmath** Yes of course, but I think he did not figure out your post because he asked for further explanations.

It was to Lotte1990 I was responding, saying essentially the same thing you did.