Question

What is the area enclosed by `r=cos(theta-(7pi)/4)+sin(-theta-(9pi)/12)` between `theta in [pi/12,(3pi)/2]`?

Answer 1

The area under the curve is 0.

Explanation:

Normally, you would use a graphing calculator and the polar integration formula to solve this problem, but there's a trick to this one.

It turns out that `cos(theta-(7pi)/4)+sin(-theta-(9pi)/12) = 0`.

We can prove this in the following way:

`sin(-theta-(9pi)/12)= -sin(theta+(3pi)/4) = -sin(theta-(5pi)/4)`

`-sin(theta-(5pi)/4)= -sin((theta-(7pi)/4)+pi/2)= -cos(theta-(7pi)/4)`

Therefore, `cos(theta-(7pi)/4)+sin(-theta-(9pi)/12)`
`= cos(theta-(7pi)/4)-cos(theta-(7pi)/4) = 0`

So the area under the curve is 0, no matter what the bounds are

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Assuming this could not be derived, you could always find the area with a graphing calculator like this:

Use the formula for polar area to solve this:

`area = 1/2int_a^br^2d theta`

`= 1/2int_(pi/12)^((3pi)/2)(cos(theta-(7pi)/4)+sin(-theta-(9pi)/12))^2d theta`

Plugging this integral into a graphing calculator gives that the area under this curve from `pi/12" to "(3pi)/2` is `0`.