Question

use integration to find the area under the graph of the function 12cos(2 theta)

Answer 1

@jcaruso What are the bounds?

Answer 2

from 0 to 2pi

Answer 3

accually they are 0 to pi/12 then you can just multiply by 8 to get area of all the petals i just dont know how to get the area

Answer 4

wait are you finding the area bound by polar curves? @jcaruso

Answer 5

yes

Answer 6

What do you mean by "area under"?

Answer 7

i mean the area enclosed by the function

Answer 8

Just integrate it haha

Answer 9

and do you know how i could get the horizontal and vertical tangents to this function

Answer 10

i found the area i just need help with the tangent lines

Answer 11

any idea? @genius12

Answer 12

Ok so the area bound by the polar curve over the given bounds is:\[\bf \frac{1}{2}\int\limits_{0}^{\pi/12}r^2 d \theta= \frac{1}{2}\int\limits_{0}^{\pi/12}144\cos^2(2\theta) d \theta=72\int\limits_{0}^{\pi/12}\frac{\cos(4 \theta)+1}{2} d \theta\]\[\bf = 36\int\limits_{0}^{\pi/12}\cos(4 \theta)+1 \ d \theta=36\left[ \frac{ \sin(4 \theta) }{ 4 } + \theta\right]_{0}^{\pi/12}=\frac{ \sqrt{3} }{ 8 }+\frac{ \pi }{ 12 }=\frac{ 3\sqrt{3}+2 \pi }{ 24 }\]

Answer 13

@jcaruso

Answer 14

@jcaruso The horizontal tangents are where the first derivative is 0; the vertical tangents are where the derivative does not exist. Note that the derivative may fail to exist when there is corners/cusps but this is often not the case with polar curves so whereever the first derivative doesn't exist will most likely be the value at which the vertical tangent occurs.