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Chapter 9: Problem 10

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

Short Answer

Expert verified
The area of the region inside the cardioid \(r=1-\sin\theta\) (above the polar axis) is \(\frac{\pi}{2}\) square units.

Step by step solution

01

Sketch the Polar Curve

Start by sketching the polar curve \(r=1-\sin \theta\). Look for patterns and try to picture it as a cardioid, knowing that a cardioid has a shape similar to a heart.
02

Determine the Boundaries of Integration

Notice that the curve intersects the polar axis (\(\theta = 0\) or \(\theta = \pi\)) when \(\sin \theta = 1\), which occurs at \(\theta = \frac{\pi}{2}\). Therefore, the interior of the curve above the polar axis is traced out as \(\theta\) ranges from \(0\) to \(\pi\). Thus, the bounds of integration are from \(0\) to \(\pi\).
03

Calculate the Integral

Now calculate the integral \(\int_0^\pi \frac{1}{2} (1-\sin\theta)^2 d\theta\). The integrand is squared because the area formula involves \(r^2\)and integrate it from \(0\) to \(\pi\).
04

Solve the Integral

After the calculation, the solution to the integral is \(\frac{\pi}{2}\).
05

Interpret the Result

It can be seen that the area of the region inside the cardioid above the polar axis is \(\frac{\pi}{2}\) square units.

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