Solve double integral

**Lotte1990** · January 29th 2013, 08:14 AM

Hello!

Can someone please tell me how to solve this double integral?

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

The following hint is given: Sketch the domain of integration.

I don't know the antiderivative of $2\sin{(y^2)}$ . I tried to guess a couple of antiderivatives, but none of their derivatives were equal to $2\sin{(y^2)}$ .

Is there something else I should do first (and how)?

Thanks!

Update: I tried switching dydx to dxdy, which results in

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

But this seems even worse, because the variable boundary is now on the outer integral.

Please help! I still can't solve this one.

**HallsofIvy** · January 29th 2013, 08:48 AM

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

**Lotte1990** · January 29th 2013, 09:03 AM

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

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

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 $2\sin{(y^2)}$ .

I need some more help, please!

**Kmath** · January 29th 2013, 11:01 AM

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 $(0,0),(\sqrt{\pi},\sqrt{\pi}),(0,\sqrt{\pi})$ .
So the domain can be alternatively determined as follows :
$0\leq y\leq \sqrt{\pi}, 0\leq x\leq y$ .
So
$\int_{0}^{\sqrt{\pi}}\int_{0}^{y}2\sin y^2 dxdy.$ Now it is straightforward.

**HallsofIvy** · January 29th 2013, 11:08 AM

Please do not stop after reading just the first sentence!
My last line was:
"Overall, y goes from 0 to $\sqrt{\pi}$ , and, for each y, x goes from x= 0 to y."

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

**Kmath** · January 29th 2013, 11:12 AM

Originally Posted by HallsofIvy

Did you not see the last line of my previous post?
"Overall, y goes from 0 to $\sqrt{\pi}$ , and, for each y, x goes from x= 0 to y."
$2\int_{y= 0}^\sqrt{\pi}\int_{x= 0}^y sin(y^2) dxdy$

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

**Lotte1990** · January 29th 2013, 11:24 AM

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

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

**Kmath** · January 29th 2013, 11:28 AM

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

**Lotte1990** · January 29th 2013, 11:40 AM

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

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

**Kmath** · January 29th 2013, 11:43 AM

Originally Posted by Lotte1990

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

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

**Lotte1990** · January 29th 2013, 11:48 AM

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

**HallsofIvy** · January 29th 2013, 02:24 PM

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.

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