Which one or ones of the following integrals produces the area that lies inside the circle r = 3 sin θ and outside the cardioid r = 1 + sin θ? (10 points)
I. 1/2 the integral from pi/6 to 5pi/6 of 9*sin^2 theta minus (1 + sin theta)^2 d theta
II. the integral from pi/ 6 to pi/2 of 9*sin^2 theta minus (1 + sin theta)^2 d theta
III. 1/2 the integral from 0 to pi/6 of 9*sin^2 theta minus (1 + sin theta)^2, d theta plus 1/2 the integral from 5pi/6 to pi of 9sin^2 theta minus (1 + sin theta)^2, d theta
A) I only
B) I and II only
C) II only
D) III only
PLEASE HELP! Use your calculator to find the length of the arc from t = 0 to t = 3 of x = 2t + 1, y = t2 - 1.
Type your answer in the space below and give 3 decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.482). (5 points)
2 answers
didn't we already do these?
∫[0,3]√((2)^2 + (2t)^2) dt = ∫[0,3] 2√(1+t^2) dt = ____
since r=3sinθ intersects r=1+sinθ at θ = π/6 and 5π/6, the area is
∫[π/6,5π/6] 1/2 (R^2-r^2) dθ
where R = 3sinθ and r = 1+cosθ
Looks like B to me, due to the symmetry of the region.
∫[0,3]√((2)^2 + (2t)^2) dt = ∫[0,3] 2√(1+t^2) dt = ____
since r=3sinθ intersects r=1+sinθ at θ = π/6 and 5π/6, the area is
∫[π/6,5π/6] 1/2 (R^2-r^2) dθ
where R = 3sinθ and r = 1+cosθ
Looks like B to me, due to the symmetry of the region.
1 answer