Question

In Exercises 3 and 4, find the area of the region bounded by the graph of the polar equation using (a) a geometric formula and (b) integration. $$ r=8 \sin \theta $$

Short answer

The area of the region bounded by the graph of the polar equation \( r = 8 \sin \theta \) is \(16 \pi\) square units.

Step-by-step solution

Solution using geometric formula

The given polar equation \( r = 8 \sin \theta \) represents a circle with radius half of the fixed number (which is 8 in this case). Hence the radius of the circle is \(r = 4\). The area \(A\) of a circle with radius \(r\) is given by the formula \(A = \pi r^2\). Substituting \(r = 4\) into the formula gives \(A = \pi (4)^2\)

Solution using integration.

The area \(A\) for a polar curve is given by the formula \( A = \frac{1}{2} \int_{a}^{b} r^2 d \theta \). For a full circle in polar form, \( a = 0, b = 2 \pi \). Substituting \( r = 8 \sin \theta \) into the formula gives \( A = \frac{1}{2} \int_{0}^{2 \pi} (8 \sin \theta)^2 d\theta \). Simplify the integral to \( 32 \int_{0}^{2 \pi} \sin^2 \theta d\theta \). Using the power-reduction identity \( \sin^2\theta = \frac{1 - \cos 2\theta }{2}\), replace \(\sin^2\theta\) and simplify.

Evaluation of integral

The integral becomes \( A = 32 \int_{0}^{2\pi} \frac{1 - \cos 2 \theta}{2} d\theta = 16 \pi - 16 \int_{0}^{2\pi} \cos 2\theta d\theta\). The second integral evaluates to zero as it involves a complete cycle of the cosine function. Thus, \(A = 16 \pi\)