Find the area inside the oval limacon r = 6 + 2 cos theta. The area inside the o

Question

Find the area inside the oval limacon r = 6 + 2 cos theta. The area inside the oval limacon is (Type an exact answer, using x as needed) Find the slope of the line tangent to the polar curve at the given point At the point where the curve intersects the origin, find the equation of the tangent line in polar coordinates. r= 8 sin theta, (4, 5 pi/6) Find the slope of the line tangent to the polar curve at the given point. At the point where the curve intersects the origin, find the equation of the tangent line in polar coordinates. r = 5-5 sin theta, (5/2, pi/6)

Explanation / Answer

Ans-4   

The limits are from = 0 to = 2 because the oval needs to make a full revolution.

The area of a polar curve is (1/2) r² d. In this case r = 6 + 2cos. Plug this in:
(1/2) (6 + 2cos)² d
Simplify by squaring the integrand.
(1/2) (36 + 24cos + 4cos²) d
The last term is difficult to integrate, so simplify it by using the identity cos² = (1 + cos2)/2. Substitute this in:
(1/2) (36 + 24cos + 4(1 + cos2)/2) d
Simplify.
(1/2) (36 + 24cos + 2 + 2cos2) d
Simplify.
(1/2) (38 + 22cos2) d
Now the integrand is much simpler. Integrate term by term. The integral of 38 is 38, and the integral of 22cos is 22sin.
(1/2)(38 + 22sin)
Simplify.
19 + 11sin
Now you can plug in the limits. Evaluate at = 2 and subtract the expression evaluated at = 0.
19(2) + 11sin(2) + (19(0) + 11sin(0))
sin(2) and sin(0) are both 0, so the last three terms become 0.

Your answer is:
38

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