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 September 25th, 2017, 12:22 AM #1 Newbie   Joined: Jan 2017 From: India Posts: 9 Thanks: 0 Math Focus: Calculus Double Integration-Finding Volume Hello, Pic:- So I was solving some questions of double integration.And I come across this que. which I didn't get. I mean I get that it is asked to find out Volume but I don't get the given data.I mean Limits explained in a tricky manner plus the answer given directly without any explaination. So can you please simplify the question for me and tell me how to solve it? Last edited by LeitHunt; September 25th, 2017 at 12:40 AM. September 25th, 2017, 10:05 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The area over which you want to integrate is the region bounded by the two parabola y= x^2 and x= y^2. There are a number of different ways to do this. I might choose to integrate with respect to y first, then with respect to x. The result must, of course, be a number so the limits of the "outer integral" must be numbers, not functions of x and y. I can see that, in order to cover that region, x must go from 0, on the right, to 1 on the left. The inner integral, with respect to y, must go from the lower parabola, y= x^2, to the upper parabola, x= y^2 or y= sqrt(x). That is, for each x, y goes from x^2 to sqrt(x): $\displaystyle \int_0^1 \int_{x^2}^{\sqrt{x}} f(x, y) dydx$. Or you could decide to integrate with respect to x first, then y. Then the "outer" integral will be with respect to y and, to cover the given region, must go from y= 0 at the bottom to y= 1 at the top. And, for each y, x must go from the left parabola, y^2= x or x= sqrt{y} to the upper parabola, x= y^2. The integral is $\displaystyle \int_0^1\int_{\sqrt{y}}^{y^2} f(x,y)dxdy$. Those are so similar because of the symmetry of the problem. Also, in going from x= y^2 to y= sqrt(x) and from y= x^2 to x= sqrt(y), we use the fact that the region of integration is in the first quadrant, so both x and y are positive to choose "sqrt(x)" and "sqrt(y)" rather than "-sqrt(x)" and "sqrt(x)". Last edited by greg1313; September 25th, 2017 at 10:37 AM. September 25th, 2017, 10:20 PM   #3
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Quote:
 Originally Posted by Country Boy The area over which you want to integrate is the region bounded by the two parabola y= x^2 and x= y^2. There are a number of different ways to do this. I might choose to integrate with respect to y first, then with respect to x. The result must, of course, be a number so the limits of the "outer integral" must be numbers, not functions of x and y. I can see that, in order to cover that region, x must go from 0, on the right, to 1 on the left. The inner integral, with respect to y, must go from the lower parabola, y= x^2, to the upper parabola, x= y^2 or y= sqrt(x). That is, for each x, y goes from x^2 to sqrt(x): $\displaystyle \int_0^1 \int_{x^2}^{\sqrt{x}} f(x, y) dydx$. Or you could decide to integrate with respect to x first, then y. Then the "outer" integral will be with respect to y and, to cover the given region, must go from y= 0 at the bottom to y= 1 at the top. And, for each y, x must go from the left parabola, y^2= x or x= sqrt{y} to the upper parabola, x= y^2. The integral is $\displaystyle \int_0^1\int_{\sqrt{y}}^{y^2} f(x,y)dxdy$. Those are so similar because of the symmetry of the problem.
Ok now I get it. Its like questions of changing the order of Integration.Thank you.
But I have one confusion in this:- the question, it says constraints "x>y^2" & "y>x^2", how this is giving us the parabola? I mean I know equation of parabola but how x>y^2 can be said x=y^2 & y>x^2 can be said y=x^2. How these Limiting region is obtained from these constraints?

Quote:
 Originally Posted by Country Boy Also, in going from x= y^2 to y= sqrt(x) and from y= x^2 to x= sqrt(y), we use the fact that the region of integration is in the first quadrant, so both x and y are positive to choose "sqrt(x)" and "sqrt(y)" rather than "-sqrt(x)" and "sqrt(x)".
And one little confusion here, did you mean going from x=y^2 to x= sqrt(y) something like that?

Last edited by LeitHunt; September 25th, 2017 at 10:26 PM. Tags double, integrationfinding, volume Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shashank dwivedi Calculus 5 April 14th, 2017 12:35 PM zollen Calculus 8 April 9th, 2017 09:51 AM Aftermath Calculus 1 May 11th, 2015 01:59 AM PedroMinsk Calculus 3 December 9th, 2010 03:59 AM jim626 Calculus 0 May 4th, 2008 05:19 PM

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