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Limit problem.

Evaluate the following double integral

"f(x,y) = x^2 + y^2

where the region R is the finite region bounded by y = x^2, 1<x<2(p.s, that is supposed to be greater than or equal to) and the x axis."

I get this. It's vertically and horizontally simple. This next part has confused me, what are the limits for x and y?

THanks.

1<x<2
0<y<x^2

Original post by SimonM

1<x<2
0<y<x^2

Doing that, I get the final answer as 6.04 and it's supposed to be 12.24

doing the dy part first you get x^6/3 then integrating that again, you get x^7/21. Where am I going wrong?

Reply 3

SimonM

You should get (x^4+x^6/6), You've lost a term essentially.

Reply 4

boromir9111

Original post by SimonM

You should get (x^4+x^6/6), You've lost a term essentially.

You've lost me now. Where has the x^4 come from?

Reply 5

SimonM

Original post by boromir9111

You've lost me now. Where has the x^4 come from?

Integrating x^2 with respect to y from 0 to x^2

Reply 6

boromir9111

Original post by SimonM

Integrating x^2 with respect to y from 0 to x^2

Oh yeah lol. I don't know why I was doing the partial integration here! Thanks mate!

Reply 7

boromir9111

This next question on this same topic I can't seem to get the same answer.

"f(x,y) = 2xy

over the region bounded by the line connecting the four points O(0,0), A(1,1), B(3,1) and C(4,0)"

I believe you can do the double integration for either side, dxdy or dydx

I did dydx. I get the limits for x 0 to 4 and y, 0 to y = x. Is this correct?

(edited 13 years ago)

Anyone?

It's not necessary to bump after only an hour or so.

Your limits aren't right; when x =4 your limits imply we can have 0<=y<=4. Sketch the area described and you'll see that when x = 4, y must be 0.

Reply 10

boromir9111

Original post by DFranklin

It's not necessary to bump after only an hour or so.

Your limits aren't right; when x =4 your limits imply we can have 0<=y<=4. Sketch the area described and you'll see that when x = 4, y must be 0.