# Double Integral and Polar, Really Need Help in the next few hours

by BenMcC
Tags: double integral, multivar calculus, polar
 P: 8 I have this problem and I cannot even begin to start it. I have to hand it in today in a few hours, and I have been stuck on it for what seems like for ever. It reads: By using polar coordinates evaluate: ∫ ∫ (2+(x^2)+(y^2))dxdy R where R={x,y}:(x^2)+(y^2)≤4,x≥0,y≥0} Hint: The region R is the quarter of a circular disc with radius r=2 that lies in the fourth quadrant. Any help would be greatly appreciated, I have no idea what else to do. Thanks
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 What are the limits on the radius and the angular variable? (By what you have written, this lies in the first quadrant, not the fourth)
 P: 8 It only states that the radius, r=2. And since it's in the fourth quadrant, the limits of the angle would be just 3∏/2 to 2∏ I believe. But I have no idea how to set up this problem
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## Double Integral and Polar, Really Need Help in the next few hours

Why not?
1. What is the area element in polar coordinates?
2. How is the integrand to be represented in polar coordinates?
 P: 8 I typed the entire problem as it is on the assignment. There's nothing else with this problem
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 Quote by BenMcC I typed the entire problem as it is on the assignment. There's nothing else with this problem
Yes.
And you ought to know, by what you have learnt, how to:
1. Replace the quadratic area element with the proper polar coordinate area element.
2. Convert the integrand into a function of the polar coordinates.
3. How to set up the new limits of the integral.
 P: 8 What I tried with this problem: R:x≥0,y≥0 R≤4, so x^2+y^2≤4 -2≤x≤2, -sqrt(4-x^2)≤y≤sqrt(4-x^2) The integral looked like 2∏ ∫ ∫ (2+r^2)dr dθ 3∏/2 R My professor does not give good notes, so I can't really follow his examples
 P: 28 arildno has given you such big hints. 1.dxdy in polar becomes rdrd(theta) 2.(2+(x^2)+(y^2)) is your surface function,simply put x=rcos(theta) and y=rsin(theta) in the function of (x,y)-because theta is measured wrt. x-axis. 3.you have to think in terms of r and (theta).Your region (which is a quarter circle on the x-y plane-1st quadrant) can be covered completely if you vary r between 0 and 2 and (theta) between 0 and pi/2. Try visualizing the situation. PS:I may be wrong.
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