Find the area bounded by y=2x^2 and y^2=4x
First you need the intersection
(2x^2)^2 = 4x
4x^4 - 4x = 0
x(x^3 - 1) = 0
x=0 or x = 1
see:
http://www.wolframalpha.com/input/?i=y%3D2x%5E2+,+y%5E2%3D4x
effective height = ?(4x) - 2x^2
= 2 x^(1/2) - 2x^2
area = ?(2 x^(1/2) - 2x^2) dx from x = 0 to 1
= [ (4/3)x^(3/2) - (2/3)x^3] from 0 to 1
= (4/3 - 2/3) - 0
= 2/3
The given curves are:
y=2x^2 and y=2x^(1/2)
The desired area lies between x=0 and x=1
The differential area is:
[2x^(1/2) - 2x^2]dx
The integral has limits:
x = 0, x = 1.
The result of integration is:
Area = 2/3.
QED
To find the area bounded by the curves y = 2x^2 and y^2 = 4x, we need to determine the points of intersection between the two curves and then use integration to calculate the area between them.
Step 1: Finding the points of intersection.
To find the points where the two curves intersect, we equate the two equations:
2x^2 = y^2 = 4x
Rearranging the equation, we get:
2x^2 - 4x = y^2
Since we have a square term (y^2), let's complete the square to simplify the equation:
2x^2 - 4x + 4 = y^2 + 4
2(x^2 - 2x + 1) = y^2 + 4
2(x - 1)^2 = y^2 + 4
Next, equate y^2 + 4 to zero to eliminate one variable:
y^2 + 4 = 0
y is imaginary here, so there are no real points of intersection for y^2 = 4x and y = 2x^2.
This means that the two curves do not intersect in the real plane, and therefore, the area bounded by them is zero.
Hence, the area bounded by y = 2x^2 and y^2 = 4x is zero.