My Math Forum Double integral 16dA, limits defined by circle

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 March 18th, 2009, 02:57 PM #1 Newbie   Joined: Mar 2009 Posts: 9 Thanks: 0 Double integral 16dA, limits defined by circle How do I solve this double integral? $\int \int_{S1} \16dA$ where S1 is the circle $(x,y)\ |\ x^2+y^2 \le 9$
 March 18th, 2009, 08:25 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,634 Thanks: 2080 Change to polar coordinates.
November 8th, 2016, 06:09 AM   #3
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Quote:
 Originally Posted by ;23464 How do I solve this double integral? $\int \int_{S1} 16dA$ where S1 is the circle $(x,y)\ |\ x^2+y^2 \le 9$
This is just $\displaystyle 16\int_S\int dA$ which is just 16A: 16 times the area of S.

What is the area of a circle with radius 3?

 November 8th, 2016, 08:55 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Frankly, I think OP is asking for more than the area of a circle. Without constant 16 $\displaystyle \int dA = \int_{0}^{2\pi}\int_{0}^{R}rdrd\theta=\int_{0}^{2\ pi}\frac{R^{2}}{2}d\theta=\pi R^{2}$ and R =3 Last edited by zylo; November 8th, 2016 at 09:19 AM. Reason: R=3

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