To find the area enclosed by the curve for

MATHEMATICS · Single correct · AP · Calculus BC · Parametric, Polar, and Vector-Valued Functions · Parametric, Polar, and Vector-Valued Functions

Problem

To find the area enclosed by the curve

r=e^{\sin\theta}

for

0\le\theta\le\pi

, one student reports

A\approx4.97

, obtained from a calculator’s numerical integration utility. A second student claims the exact value is

A=\tfrac{\pi}{2}\left(e-\dfrac1e\right)

. Which statement best evaluates these two results? (a) Both students are correct, since

4.97

is simply a rounded version of

\dfrac{\pi}{2}\!\left(e-\dfrac1e\right)

. (b) Neither student is correct; the area should be

\displaystyle A=\int_{0}^{\pi} e^{\sin\theta}\,d\theta\approx2.48

. (c) The first student is more plausible; the defining integral

\displaystyle \tfrac12\int_{0}^{\pi}e^{2\sin\theta}\,d\theta

has no elementary antiderivative, so a numerical result such as

4.97

is appropriate. (d) The second student is more likely correct because the function

e^{\sin\theta}

is symmetric on

[0,\pi]

, making the area a simple multiple of

\pi

, namely

\dfrac{\pi}{2}\!\left(e-\dfrac1e\right)