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March 29, 2015

Posted by **Tracy** on Tuesday, November 22, 2011 at 2:33pm.

r=(2 sin theta)+(4 cos theta)

A. (sqrt2)(pi)

B. (sqrt5)(pi)

C. (2)(pi)

D. (5)(pi)

- Calculus still confused -
**drwls**, Tuesday, November 22, 2011 at 3:06pmI am confused also. That does not look to me like the polar equation for a circle.

- Calculus still confused -
**Steve**, Tuesday, November 22, 2011 at 11:15pmIt is the equation of a circle.

r = 2sinθ + 4cosθ

r*r = 2rsinθ + 4rcosθ

x^2 + y^2 = 2y + 4x

(x^2 - 4x + 4) + (y^2 - 2y + 1) = 4+1

(x-2)^2 + (y-1)^2 = 5

However, we need to find the area using polar coordinates:

As I noted in my earlier posting, the first quadrant means 0 <= θ <= pi/2

So, we integrate

A = 1/2 Int(r^2 dθ)[0,pi/2]

= 1/2 Int(4sin^2θ + 16sinθ cosθ + 16 cos^2θ dθ)[0,pi/2]

Recalling some trig identities:

sin^2θ + cos^2θ = 1

sin 2θ = 2sinθ cosθ

cos^2θ = (1 + cos2θ)/2

A = 1/2 Int(4 + 8sin2θ + 6(1 + cos2θ) dθ)[0,pi/2]

= 1/2 Int(10 + 8sin2θ + 6cos2θ)[0,pi/2]

= 1/2(10θ - 4cos2θ + 3sin2θ)[0,pi/2]

= (5θ - 2cos2θ + 3/2 sin2θ)[0,pi/2]

= (5*pi/2 - 2(-1) + 0) - (0 - 2(1) + 0)

= 5pi/2 + 4

= 11.854

I don't get any of the given choices. And, my answer agrees with wolfram dot com:

integrate .5*((2 sin theta)+(4 cos theta) )^2 dtheta, theta=0..pi/2

Take off the limits of integration to see that their indefinite integral also agrees with mine.

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