# Area of the region inside the unit circle

## Homework Statement

The area in the region inside the unit circle and above the graph of f(x) = x^5

## Homework Equations

I don't know how to type the equation in here but the area is the integral between two integration points of the higher curve minus the lower curve.

## The Attempt at a Solution

I tried using the integration points 0 and 2.02381 (the farthest the graph will stretch out according to derive), but that's the area outside and under the curve, i think the question is asking for the area INSIDE the curve.
As far as I know you can't do it with respect to y because it's infinite.

Thanks.

## Answers and Replies

I'm not sure why you'd be using those integration points. For a unit circle, what is the largest possible value of x and/or y? Or, maybe think about the equation for a unit circle. Where does f(x) cross the unit circle?

(1,1)? So it'll be from 0 to 1 correct?

If we were to do it that way, the point of intersection would be 1 and i'm guessing the radius will be on top of the curve?
So it'll be... SQRT(1-x^2) - (x^5)?

Mark44
Mentor
(1, 1) is not a point on the circle.

Finding the points of intersection entails solving the equation sqrt(1 - x^2) = x^5. If you square both sides, you get a 10th degree equation that has no easy-to-get solutions.