- #1

- 15

- 0

I need help setting up this problem:

Use a double integral to find the area of the region enclosed by the curve

r=4+3 cos (theta)

Thanks

- Thread starter PhysicsMajor
- Start date

- #1

- 15

- 0

I need help setting up this problem:

Use a double integral to find the area of the region enclosed by the curve

r=4+3 cos (theta)

Thanks

- #2

- 17

- 0

1)You'll have to draw a polar curve to help you out with this question. From the drawn polar curve, you'll get the minimum value of r to be 1. (when cos(theta) is negative)

2) Thus it follows that the range for r is 1<=r<=4+3cos(theta)

Hence we'll integrate r from 1 to 4+3cos(theta)

3) For the range of theta, you can use the range from 0 to pi for simplicity in calculations.(Just multiply the answer by 2 to obtain the full area.)

I hope this helps =)

- #3

- 13,016

- 566

Yeah,a plot might help u convince why the limit wrt [itex] \theta [/itex] need to be 0 and [itex] \pi [/itex] and why you shouldn't integrate from 0 to [itex] 2\pi [/itex]

I think it's a cardioide.

Daniel.

I think it's a cardioide.

Daniel.

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