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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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 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.

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