Finding limits for a double integral?

I don't think there's a "mechanical" way to do it. I have to graph the region, then figure out how that same region can be delineated in the other coordinates.

Here, the region of integration is a right triangle with its right angle at (0,6), with equal legs of length 6 and hypotenuse along the line y=x. That very strongly suggests that θ goes from π/4 to π/2, and that we should do r-integration first.

If you draw in an element of polar area, in this case it always starts at the origin and terminates at the line y = 6. Since y = r sin(θ), y = 6 is equivalent to r sin(θ) = 6, or r = 6 csc(θ)

So the lower limit is r = 0, and the upper limit is r = 6 csc(θ)

Finally, the integrand (which I suspect you know how to do): x becomes r cos(θ) and dx dy becomes r dr dθ

Double Integral Limits

double integrals are particularly uncomplicated as quickly as you artwork out the way you stumble on the obstacles. As a familiar rule what you're doing is figuring our the section in the xy aircraft. a million. continually attempt to attraction to the two bounds. first inequality exhibits that x is going from 0 to y and the 2d exhibits that y is going from 0 to a million. in case you graph it out it sounds like this / a slant. 2. to set this up, the 1st inequality is in terms of x and the 2d interms of y. Which ever certain is given by potential of in simple terms numbers is the final bit you combine. so which you desire to eliminate the greater variable. 3. so this may be (int 0 to a million)(int 0 to y) 2 dxdy what happens now's that once you combine the interior fundamental, you get 2x evaluated at y and 0 for the reason that x =y then the assessment provides 2(y)-2(0) that's in simple terms 2y. Now the outer integrals is in terms of the shrink of integration of y. combine and evaluate. one thank you to think of approximately it is: (fundamental bounded by potential of y=0 to y=a million)(fundamental bounded by potential of x=0 to x=y)dxdy i desire this enables.