Area enclosed by 2 graphs

Question

Find the area A of the region between the graphs of the equations

y² = 8x

x² = 8y

Answer 1

The upper (blue) curve is y = sqrt(8x) and the lower curve is y = x²/8. To find the area enclosed, integrate the two curves and subtract them.

They intersect at (0,0) and at (8,8) so integrate between 0 and 8.

Integral from 0 to 8 of sqrt(8x)dx = 128/3 (antiderivative = (4√2/3)x^3/2) and the Integral from 0 to 8 of x²/8 = 64/3 (antiderivative = x³/24) so the area between the curves is 128/3 - 64/3 = 64/3

Answer 2

Hi Carly!

I tried to record a video explaining the problem, but unfortunately there was an error, and it won't let me! I've typed up an explanation below and also written up a solution for you to follow: https://imgur.com/a/3rwx7y0

1) So first off, looking at the problem, we're looking for the area of a region bounded by those two functions. That tells me that we're probably going to integrate something.

Looking at the two equations, I see that one of them is a y² = ... and the other is x² = ...

I'm going to manipulate both equations to get them into the form y = ...

For y² = 8x, this would involve taking the square root of both sides, which gives me y = √(8x). Technically, there should be a plus or minus, but we'll soon see why that doesn't matter for this problem.

For x² = 8y, I'm going to divide both sides by 8, giving me y = x²/8

2) Now if I do a rough sketch of these graphs, I see that y = x²/8 is below y = √(8x). Setting the two equations equal to each other, I get x²/8 = √(8x). I can solve this equation to see where the two graphs intersect, which is at x = 0 and x = 8. These are the boundaries of my integration.

3) For the actual integral, I set it up as the integral of √(8x) - x²/8 from 0 to 8. I used u-substitution to solve part of the integral, but that part is easier to see than explain, so please take a look at the linked image.

And once you integrate, you're done! Hope this was helpful!