Consider the closed curve between 0 <= theta < 2pi given by r(theta) = 6 + alpha sin theta, where alpha is some real constant strictly between 0 and 6. The area in this closed curve is 97pi/2. Calculate the value of alpha.
Student uses the definition of area [A = 1/2 integral r(theta)^2 d theta], and proceeds using standard integration techniques to give a quadratic solvable for alpha. [alpha^2 = 25] Thus, alpha = 5.
GC
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