Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I’m really having problems getting suitable limits when using double integration. For example:

Let E be the region defined by {(x,y): y-x <= 2, x + y >= 4, 2x + y <= 8} Sketch the region E.

The sketching is never a problem. So for this one we have:

y = 8-2x, y=x+2 and y=4-x - the triangle between all the lines is the region E.

Find the area of E and also find

Integral Integral E 1/x dy dx

The integration isn’t the problem - the limits are. I’m really confused as to how to get the correct limits. Our notes say to fix a variable and then look at the boundary of the other. So if we fix x in (1,4) then taking y=x+2 and y=8-2x will give me a greater area then needed....

I hope someone can explain this to me as I’m rather confused!

Here is a similar problem:

I = Integral x=0, 1 Integral y=0, x ( x^3 / (x^2 + y^2)1.5 ) dy dx

Sketch the region (this is fine). Consider x=rcos(theta) and y=rsin(theta). Find the Jacobian (again this was fine) - its came out as r. What is the image of R under the transformation? (not sure how to find/sketch this)/ Use the transformation to evaluate the integral I.

Making the substitution gave cos(theta)^3/r but idk where to go from there....

share|cite|improve this question
@user4645: You've been ignoring my comments here:… and here:…. Please show more respect for the site. If you go on like this I may start flagging your questions. – joriki May 27 '11 at 16:27
@joriki: I posted an answer before I saw your comment. I think your comment is a good one, so deleted the answer. Let me think if you think I should repost. – Ross Millikan May 27 '11 at 16:43
@Ross: Thanks. My tendency would be to wait with reposting until user4645 has made some effort to cooperate. – joriki May 27 '11 at 16:48
@user4645: Find the corners of the triangle. I get $(1,3)$, $(3,5)$, and $(4,0)$. We have to decide whether to integrate first w.r.t. $x$ or w.r.t. $y$. Doing $x$ first sounds hard, you will get logs, which you need to later integrate. So try to integrate first w.r.t. $y$. The "top" boundary changes. So break up the region using a vertical line through $(3,5)$. In the left region, $y$ goes from $4-x$ to $x+2$. Integrate (trivial). Then integrate w.r.t. $x$, from $1$ to $3$. Please leave a message if the process is not clear. – André Nicolas May 28 '11 at 4:10
hi... ive corrected my mistakes. sorry about that! – user4645 May 30 '11 at 17:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.