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

 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
 Sci Advisor HW Helper PF Gold P: 12,016 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
 Sci Advisor HW Helper PF Gold P: 12,016 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? 3. The maximal radius equals 2. What is the minimal radius?
 P: 8 I typed the entire problem as it is on the assignment. There's nothing else with this problem
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PF Gold
P: 12,016
 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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PF Gold
P: 12,016
 Quote by BenMcC 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
Remember that the area element is r*drd(theta).

The integral is radially symmetric, so it doesn't really matter if you integrate in the fourth quadrant or the first. (But the inequalities you gave for x and y denotes the first, not the fourth)

And, the limit on r is from 0 to 2
 P: 8 I ended up working it out and got an answer of 8pi+1/2. Not entirely confident in that answer
 Sci Advisor HW Helper PF Gold P: 12,016 What was your integrand? Notice that the integration with respect to the angular variable gives you the factor pi/2 This is then multiplied with your result from the r-integration.

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