How do you sketch the region enclosed by #y=x^2, y^2=x# and find the area?

Question

How do you sketch the region enclosed by  #y=x^2, y^2=x# and find the area?

Ezra Adams · Accepted Answer

#1/3# areal units.

The common points are (0, 0) and (1, 1). #y^2 = x>=0 and x^2=y>=0#. So, the petal-like common area is between O and (1, 1) Area = #int# (y-difference) dx, with x from 0 to 1. #= int (sqrtx-x^2) dx#,with x from 0 to 1. #=[2/3x^(3/2)-x^3/3]#, between x = 0 and 1. #=1/3, areal units. graph{(y-x^2)(y^2-x)=0 [-2.51, 2.51, -1.255, 1.255]}

Christopher Allers · Answer

undefined To sketch the region enclosed by $y = x^2$ and $y^2 = x$, you first need to find the points of intersection between the two curves. By solving the system of equations formed by equating $y = x^2$ and $y^2 = x$, you can find the $x$-coordinates of the intersection points. Once you have the $x$-coordinates, you can determine the corresponding $y$-coordinates by substituting them into either equation.

Next, plot these points on a coordinate plane and sketch the curves $y = x^2$ and $y^2 = x$. The region enclosed by these curves will be the area between the points of intersection.

To find the area of the enclosed region, you need to set up a definite integral. Since the region is bounded by the curves, the integral will be the difference between the upper and lower curves with respect to $x$. The bounds of integration will be the $x$-coordinates of the points of intersection.

After setting up the integral, evaluate it to find the area of the enclosed region. This can be done by finding the antiderivatives of the functions involved and applying the Fundamental Theorem of Calculus, or by using other integration techniques.

Once you have the definite integral set up and evaluated, the result will give you the area of the region enclosed by $y = x^2$ and $y^2 = x$.