Do I use a double integral? Thanks.
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Do what the hint says and find where r =0
This will tell you what theta values are the bounds for the inner loop.
@ = theta
0 =1 -2cos(@)
0.5 = cos(@)
@ = pi/3, 5pi/3
Do a test for a @ = pi/2:
1- 2cos(pi/2) = 1
We see that from pi/3 to 5pi/3, the radius is positive, so this must be the big loop.
Check by doing another test for @ =0:
1-2cos(0) = -1
Yep, the inner loop is from -pi/3 to pi/3
Now it's just integration.
Integrate r with respect to @:
indefinite integral of r = @ - 2sin@ + C
I assume you know how to do simple integration, remember d/dx sin(x) = cos(x)
Call the indefinite integral: R(@) = @ - 2sin@ + C
Area of big loop = |R(5pi/3) - R(pi/3)| = ~7.6539
Area of small loop = |R(pi/3) - R(-pi/3)| = ~1.3697
Note the absolute value is taken because we are interested in the positive area, and the negative radius of the small loop will yield a negative area.
Subtract the small loop from the big loop to get the desired area of ~6.2842.
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Verified answer
Do what the hint says and find where r =0
This will tell you what theta values are the bounds for the inner loop.
@ = theta
0 =1 -2cos(@)
0.5 = cos(@)
@ = pi/3, 5pi/3
Do a test for a @ = pi/2:
1- 2cos(pi/2) = 1
We see that from pi/3 to 5pi/3, the radius is positive, so this must be the big loop.
Check by doing another test for @ =0:
1-2cos(0) = -1
Yep, the inner loop is from -pi/3 to pi/3
Now it's just integration.
Integrate r with respect to @:
indefinite integral of r = @ - 2sin@ + C
I assume you know how to do simple integration, remember d/dx sin(x) = cos(x)
Call the indefinite integral: R(@) = @ - 2sin@ + C
Area of big loop = |R(5pi/3) - R(pi/3)| = ~7.6539
Area of small loop = |R(pi/3) - R(-pi/3)| = ~1.3697
Note the absolute value is taken because we are interested in the positive area, and the negative radius of the small loop will yield a negative area.
Subtract the small loop from the big loop to get the desired area of ~6.2842.