Evaluate the double integral

[boneill3]

Homework Statement

\int \int_R (x+y) dA R is the region y = x^2 and y = x^(1/2)

Homework Equations

I've sketched the graph and the functions are equal at (0,0) and (1,1)

The Attempt at a Solution

Would the limits of the integral be:

\int_{0}^{1) \int_{0}^{1} (x+y) dydx


regards

 

[boneill3]

Sorry would it be


Would the limits of the integral be:

\int_{0}^{1) \int_{x^1/2}^{x^2} (x+y) dydx


regards

 

[djeitnstine]

boneill3, double integration is not like single variable integration. You need to always express your first limits in terms of your second integrating variable. Perhaps some math guru can come in here and give you a more rigorous definition.

But here your limits are \int_{0}^{1) \int_{x^2}^{x^(1/2)} dydx

perhaps here may help:

http://ltcconline.net/greenl/courses/202/multipleIntegration/iteratedIntegrals.htm

edit: I was a little too slow on the reply

 

[Feldoh]

boneill3, double integration is not like single variable integration. You need to always express your first limits in terms of your second integrating variable.

Not always, sometimes the two are independent, it's just in this case they are not.

 

[djeitnstine]

Thanks, I just know how to do it, not the reasoning behind it =]

 

[boneill3]

Thank you for that web link It explains it a lot better than my lecture notes.

I see that I got the functions upper and lower boundsupside down as well.

\int_{x^2}^{x^1/2}

instead of

\int_{0}^{1) \int_{x^1/2}^{x^2} (x+y) dydx

 

[boneill3]

Hopefully I've got this one right

\int \int_R (xcosy) dA R in the region y = x and y = 0 and x = \pi

This is a triangular region bounded on the right by x = \pi, below by y = 0 and above by y = x.

So the limits are:
0<= x <= \pi and 0 <=y <= x

\int_{0}^{\pi) \int_{0}^{x} (xcosy) dydx

regards
Brendan

 

[Feldoh]

It looks right

 

[Enzo]

Hopefully I've got this one right

\int \int_R (xcosy) dA R in the region y = x and y = 0 and x = \pi

This is a triangular region bounded on the right by x = \pi, below by y = 0 and above by y = x.

So the limits are:
0<= x <= \pi and 0 <=y <= x

\int_{0}^{\pi) \int_{0}^{x} (xcosy) dydx

regards
Brendan

That's right. The limits of integration are presented through the analytical expression of the domain:

So the domain of this question would be:

D = {(x,y)| 0<=y<=x, 0<=x<=pi }

 

[boneill3]

Thanks guys