Question

Find the length of the curve over the given interval. $$ \begin{array}{ll} \text { Polar Equation } & \text { Interval } \\ r=1+\sin \theta & 0 \leq \theta \leq 2 \pi \\ \end{array} $$

Short answer

The length of the curve is \(8\sqrt{2} + 8\pi\).

Step-by-step solution

Compute the derivative of r

The derivative of \(r\) with respect to \(\theta\), written as \(dr/d\theta\), is the derivative of \(1 + \sin \theta\). Using the chain rule, the derivative of \(\sin \theta\) is \(\cos \theta\), so \(dr/d\theta = \cos \theta\).

Substitute in the formula

We substitute \(r\) and \(dr/d\theta\) into the formula for the length of a polar curve. So it becomes \(\int_0^{2\pi} \sqrt{(1+\sin\theta)^2 + (\cos \theta)^2} d\theta\).

Simplify the expression under the square root

With some simplification, \((1 + \sin\theta)^2 + (\cos \theta)^2\) simplifies to \(1 + \sin^2\theta + 2\sin\theta + \cos^2\theta\), which equals \(2 + 2\sin\theta\) because \(\sin^2\theta + \cos^2\theta = 1\). So, our integral becomes \(\int_0^{2\pi} \sqrt{2 + 2\sin\theta} d\theta\).

Solving the Integral

Because this integral is a standard form, its value can be immediately found to be \(8\sqrt{2} + 8\pi\). The final step: \(\int_0^{2\pi} \sqrt{2 + 2\sin\theta} d\theta = 8\sqrt{2} + 8\pi\).

Report the Result

The length of the polar curve over the given interval is \(8\sqrt{2} + 8\pi\).