• Support PF! Buy your school textbooks, materials and every day products Here!

Evaluate the double integral

  • Thread starter boneill3
  • Start date
  • #1
127
0

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
 

Answers and Replies

  • #2
127
0
Sorry would it be


Would the limits of the integral be:

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


regards
 
  • #3
djeitnstine
Gold Member
614
0
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
 
  • #4
1,341
3
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.
 
  • #5
djeitnstine
Gold Member
614
0
Thanks, I just know how to do it, not the reasoning behind it =]
 
  • #6
127
0
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
 
  • #7
127
0
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
 
  • #8
1,341
3
It looks right
 
  • #9
17
0
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 }
 
  • #10
127
0
Thanks guys
 

Related Threads for: Evaluate the double integral

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
949
  • Last Post
Replies
4
Views
772
  • Last Post
Replies
18
Views
714
Replies
4
Views
775
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
2
Views
847
  • Last Post
Replies
16
Views
2K
Top