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October 1st, 2017, 11:18 AM   #1
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Question on Multiple Integration

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.
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October 1st, 2017, 02:38 PM   #2
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Plot the limits of y on x-y graph. Look at what the limits would be if you did x first.
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October 1st, 2017, 05:36 PM   #3
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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 xy-coordinate 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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