Question

Find the area of the shaded region. r = 1 + cos ? A coordinate plane is given. A cardioid, a circle, and a shaded region are graphed. - The cardioid r = 1 + cos(?) has a cusp on the left side of the graph at (0, 0), then goes up and left, changes directions, goes up and right, crosses the y-axis, changes directions where it crosses the circle, goes down and right, crosses the x-axis, goes down and left, changes directions where it crosses the circle, goes up and left, crosses the y-axis, changes directions, goes up and right, and ends at the starting point. - The circle r = 3 cos(?) has its center on the positive horizontal axis and its left side passes through the origin. - The shaded region is the area inside the cardioid and outside the circle above the horizontal axis.

          Find the area of the shaded region.
r = 1 + cos ?
A coordinate plane is given. A cardioid, a circle, and a shaded region are graphed.
- The cardioid r = 1 + cos(?) has a cusp on the left side of the graph at (0, 0), then goes up and left, changes directions, goes up and right, crosses the y-axis, changes directions where it crosses the circle, goes down and right, crosses the x-axis, goes down and left, changes directions where it crosses the circle, goes up and left, crosses the y-axis, changes directions, goes up and right, and ends at the starting point.
- The circle r = 3 cos(?) has its center on the positive horizontal axis and its left side passes through the origin.
- The shaded region is the area inside the cardioid and outside the circle above the horizontal axis.

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Find the area of the shaded region.
r = 1 + cos ?
A coordinate plane is given. A cardioid, a circle, and a shaded region are graphed.
- The cardioid r = 1 + cos(?) has a cusp on the left side of the graph at (0, 0), then goes up and left, changes directions, goes up and right, crosses the y-axis, changes directions where it crosses the circle, goes down and right, crosses the x-axis, goes down and left, changes directions where it crosses the circle, goes up and left, crosses the y-axis, changes directions, goes up and right, and ends at the starting point.
- The circle r = 3 cos(?) has its center on the positive horizontal axis and its left side passes through the origin.
- The shaded region is the area inside the cardioid and outside the circle above the horizontal axis.

Added by Margaret W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supreeta N

12/04/2023

Step 1

To do this, we set their equations equal to each other: $$r = \cos(\theta)$$ Since the circle has the equation $r = \cos(\theta)$, we can substitute this into the cardioid equation: $$\cos(\theta) = \cos(\theta)$$ This equation is true for all values of Show more…

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Find the area of the shaded region. The coordinate plane is given. A cardioid, circle, and shaded region are graphed. The cardioid r = 1 + cos(̘) has a cusp on the left side of the graph at (0, 0), then goes up and left, changes directions, goes up and right, crosses the y-axis, changes directions where it crosses the circle, goes down and right, crosses the x-axis, goes down and left, changes directions where it crosses the circle, goes up and left, crosses the y-axis, changes directions, goes up and right, and ends at the starting point. The circle r = 3 cos(̘) has its center on the positive horizontal axis and its left side passes through the origin. The shaded region is the area inside the cardioid and outside the circle above the horizontal axis.

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Transcript

00:01 Hi, here for the given question we are given that r is equal to 1 plus cos theta and r is equal to 3 cos theta.

00:08 So here in our case, this is the graph.

00:10 Now we need to calculate the area.

00:12 So first of all, we will find the limit.

00:13 So we have 1 plus cos theta equals to 3 cos theta.

00:17 So we have 1 is equal to 2 cos theta and here we have cos theta equals to 1 by 2.

00:24 So theta will be equal to pi by 3.

00:26 So here area can be calculated as integration over 0 to pi and integration over 0 to r, which is equal to 3 cos theta and r dr d theta over here.

00:40 So simplifying further, here it is limit pi by 3.

00:44 Now area can be written as 2 times of integration over 0 to here it is from 1 plus cos theta.

00:52 So 2 times integration over 0 to pi by 3.

00:56 Here it is from minus pi by 3 to pi by 3 because of the limit.

01:00 So 2 times here we have 1 plus cos theta and here 3 cos theta r dr d theta.

01:07 Now further integrating with respect to d theta, first of all, so area equals to 2 times integration over 0 to pi by 3 and here r will be as r square upon 2 and here we have 1 plus cos theta 3 times of cos theta and here we have d theta.

01:25 So substituting the value of the limit, we have integration over 0 to pi by 3.

01:29 Here it is 2 times 1 by 2 and here we have 9 cos square theta minus 1 plus cos theta whole square d theta.

01:38 Now we know that if we simplify this bracket and solve further, we have area equals to integration over 0 to pi by 3 9 times cos square theta.

01:48 It will be 8 times cos square theta minus 2 cos theta minus 1 d theta...

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