October 1st, 2017, 11:18 AM  #1 
Newbie Joined: Oct 2017 From: India Posts: 1 Thanks: 0  Question on Multiple Integration
Hiiiee!! I'mma. Jyoti,, from India. . .,, Pursuing Btech I T. 1st year Having some problems with my subjects,,, I'll be glad if you help me out with this question. I'd attached the picture of the question as an attachment. 
October 1st, 2017, 02:38 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,434 Thanks: 562 
Plot the limits of y on xy graph. Look at what the limits would be if you did x first.

October 1st, 2017, 05:36 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,940 Thanks: 794 
The integral as given takes x from 0 to 1 and, for each x, y from x to $\displaystyle \sqrt{2 x^2}$ (and it should be "dydx" not "dxdy"). In an xycoordinate system, as mathman suggests, $x= 0$ and $x= 1$ are vertical lines, $y= x$ is a slant line, though (0, 0) and (1, 1), and $\displaystyle y= \sqrt{2 x^2}$, which is the same as $\displaystyle y^2= 2 x^2$ or $\displaystyle x^2+ y^2= 2$ is the circle with center at (0, 0) and radius $\displaystyle \sqrt{2}$. That circle intersects the line $y= x$ at (1, 1) and the line $x= 0$ at $\displaystyle (0, \sqrt{2})$. So $y$, overall, goes from 0 to $\displaystyle \sqrt{2}$ and, for each $y$, with $\displaystyle 0\le y\le 1$, $x$ goes from 0 to $y$, while with $\displaystyle 1\le y\le \sqrt{2}$, $x$ goes from 0 to $\displaystyle \sqrt{2 y^2}$. The integral, with order of integration reversed, is $\displaystyle \int_0^1\int_0^{\sqrt{2 y^2}} \frac{x dxdy}{\sqrt{x^2+ y^2}}$. Do you see why that is easier to integrate than the original form? Last edited by skipjack; October 1st, 2017 at 07:20 PM. 

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