# Math

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This image has been Flagged as inappropriate Click to unflag You need to evaluate the area enclosed by the loop `r(theta) = 3 + 6sin theta` , hence, you need to evaluate the following definite integral, such that:

`int_(theta 1)^(theta 2) r(theta)d theta`

Since the problem does not provide the limits of integration `theta 1 ` and `theta 2` , you only may evaluate the integral such that:

`int_(theta 1)^(theta 2) r(theta)d theta = int_(theta 1)^(theta 2) (3 + 6sin theta)d theta`

Using the property of linearity of integral yields:

`int_(theta 1)^(theta 2) r(theta)d theta = int_(theta 1)^(theta 2) 3 d theta + int_(theta 1)^(theta 2) 6 sin theta d theta`

`int_(theta 1)^(theta 2) r(theta)d theta = (3 theta - 6 cos theta)|_(theta 1)^(theta 2)`

Using the fundamental theorem of calculus, yields:

`int_(theta 1)^(theta 2) r(theta)d theta = 3 theta 1 - 6 cos theta 1 - 3 theta 2 + 6 cos theta 2`

`int_(theta 1)^(theta 2) r(theta)d theta = 3 theta 2 - 3 theta 1`

`int_(theta 1)^(theta 2) r(theta)d theta = 3(theta 2 - theta 1)`

Hence, evaluating the area enclosed by the given inner loop yields `int_(theta 1)^(theta 2) r(theta)d theta = 3(theta 2 - theta 1).`

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