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Chapter 9: Problem 9

Find the area of the region. Interior of \(r=1-\sin \theta\)

Short Answer

Expert verified
The area of the region inside the polar curve \(r=1-\sin \theta\) is \(\frac{3\pi}{2}\).

Step by step solution

01

Set up the integral to find the area

The formula to find the area \(A\) in polar coordinates is given by \(A = \frac{1}{2} \int r^2d\theta \). Plugging the given polar equation \(r= 1- \sin \theta \) into the area formula, the integral becomes \(A = \frac{1}{2} \int_0^{2\pi} (1 - \sin \theta)^2d\theta.\)
02

Simplify integral

Expand the integral \(A = \frac{1}{2} \int_0^{2\pi} (1 - 2\sin \theta + sin^2 \theta) d\theta\). The integral can be split into three separate integrals. So, \(A= \frac{1}{2} [\int_0^{2\pi} d\theta - 2\int_0^{2\pi} \sin \theta d\theta + \int_0^{2\pi} \sin^2 \theta d\theta].\)
03

Evaluate the integrals

Now evaluate each integral separately. The first integral evaluates to \(2\pi\). The second integral evaluates to 0 as the integral of \(\sin \theta\) over the interval 0 to \(2\pi\) is 0. To calculate the third integral, remember that \(\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}\). Using this transformation, the third integral evaluates to \(\pi\).
04

Compute the total area

The total area under the polar curve is the sum of the separate integrals which is \(A=\frac{1}{2}(2\pi+0 + \pi)= \frac{3\pi}{2}\).

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Most popular questions from this chapter

Determine whether \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal parallel, or neither. $$ \begin{array}{l} \mathbf{u}=-\frac{1}{3}(\mathbf{i}-2 \mathbf{j}) \\ \mathbf{v}=2 \mathbf{i}-4 \mathbf{j} \end{array} $$

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In Exercises 29 and 30 . find the direction angles of the vector. $$ \mathbf{u}=3 \mathbf{i}+2 \mathbf{j}-2 \mathbf{k} $$
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