Posted by Log on Saturday, March 21, 2009 at 8:17pm.
You have to integrate 1/2 r^2 from theta = 0 to 2 pi.
1/2 r^2 = 1/2 [25 + 20 sin(theta)
+ 4 sin^2(theta)]
The sin(theta) term doesn't contribute to the integral and sin^2(theta) can be replaced by 1/2 [because it would yield the same contribution as cos^2(theta) and together they are equal to 1].
So, the area is given by:
pi[25 + 2 ] = 27 pi
Related Questions
Calculus - Find the area of the region enclosed between y = 2sin (x) and y = ...
Calculus - Find the area of the region enclosed by the given curves: y=e^6x, y=...
Calc 2: Area under the curve - Find the area of the region enclosed between y=...
Calculus - Let f(x)=x^3+3x and g(x=5x^2-x Determine the area of the region ...
pre-calc - area of a rectangular region: a farmer wishes to create two ...
Calculus - please help! - Suppose that 0 < c < ¥ð/2. For ...
Calculus - Calculate the area of the common interior of r = 2sin(theta) and r = ...
calculus - find the area of the region bounded by the polar curve r=sqrt(6ln(...
calculus - the region enclosed by 2y=5xy=5 and 2y+3x=8. Decide whether to ...
calculus - Sketch the region enclosed by x+y^2=42 and x+y=0. Decide whether to ...
For Further Reading