# Calculate the area above y=x^2 and to the right of x=y^2 using integrals.

## y = x^2 is 0 at x = 0, .25 at x = 1/2 and 1 at x = 1

y = +/- sqrt x is 0 at x=0, .707 at x = 1/2 and 1 at x = 1

sketch that and you will see that what you want is the integral from x = 0 to x = 1 of x^(1/2)dx minus the integral over the same interval of x^2 dx

## The two cuves intersect at (x=0, y=0) and (x=1, y=1). The area between them is

(Integral of) x^1/2 - x^2 dx

0->1

Evaluate x^(3/2)/(3/2) - x^3/3.

at the two limits and take the difference

## To calculate the area above y = x^2 and to the right of x = y^2, we need to set up a proper integral.

First, let's find the points of intersection between the two curves. To do this, we set x = y^2 equal to y = x^2 and solve for x:

y^2 = x^2

Taking the square root of both sides, we get:

y = ±x

So, the points of intersection are at (x, y) = (x, x) and (x, y) = (x, -x).

Next, let's determine the boundaries of integration. The region is bounded by the curves y = x^2 and x = y^2. Since we are integrating with respect to y, we need to express these curves in terms of y.

The curve y = x^2 can be rewritten as x = √y.

The curve x = y^2 can be rewritten as y = ±√x.

Since we want the region to the right of x = y^2, we consider y = √x.

Now, we have the following boundaries of integration:

x = √y (lower bound)

x = √x (upper bound)

From these bounds, we can determine the integral setup. We integrate with respect to y because the region is vertical.

The integral for the area becomes:

A = ∫[√y, √x] (√x - √y) dy

To solve this integral, we need to find x in terms of y. From the upper bound, we solve √x = √y to get x = y.

A = ∫[√y, y] (√x - √y) dy

Next, we evaluate this integral:

A = ∫[√y, y] (y - √y) dy

To solve this integral, we expand and simplify:

A = ∫[√y, y] y dy - ∫[√y, y] √y dy

The first integral becomes:

A = [1/2 * y^2] evaluated from √y to y

A = 1/2 * (y^2 - (√y)^2)

Simplifying, we have:

A = 1/2 * (y^2 - y)

To find the final value of A, we subtract the areas below the x-axis by taking only the positive portion:

A = 1/2 * (y^2 - y) evaluated from 0 to 1

A = 1/2 * (1 - 1/2)

A = 1/4 square units

Therefore, the area above y = x^2 and to the right of x = y^2 is 1/4 square units, as calculated using integrals.