# Thread: Can someone check this - Area between curves

1. ## Can someone check this - Area between curves

Find the area between $y = 20-x^2$ and $y=x^2-12$

I found the integral to be

$\int_{-4}^{4}[(20-x^2)-(x^2-12)]dx$

and after evaluating that i found the area to be 170.66667

Does this seem correct?

Thank you?

2. ## better if along y-axis

Originally Posted by mybrohshi5
Find the area between $y = 20-x^2$ and $y=x^2-12$

I found the integral to be

$\int_{-4}^{4}[(20-x^2)-(x^2-12)]dx$

and after evaluating that i found the area to be 170.66667

Does this seem correct?

Thank you?
y=x^2-12 is a parabola pointing upwards with vertex at (0,-12) while the other is a parabola pointing downwards with vertex at (0,20). both of them meet at(+-4,4).

Now integrating along the x-axis will lead to complications. better will be to integrate along y-axis.then the integral will be from 20 to -12and the expression will be as in the attachment.

3. I just tried what you attached and i got 0 for my answer.

Does that seem right?

Thanks for the help

4. Looking at my original graph again of y=20-x^2 and y=x^2-12 i notice that there is more area above the x axis than below so it doesnt seem right to me that the area is 0. Just from looking at the graph it looks like the area should maybe be a positive number????

5. ## unable to find the mistake

Originally Posted by mybrohshi5
Looking at my original graph again of y=20-x^2 and y=x^2-12 i notice that there is more area above the x axis than below so it doesnt seem right to me that the area is 0. Just from looking at the graph it looks like the area should maybe be a positive number????
you are right. the answer comes out to be 0.which is wrong. i am not able to find out my mistake. either there is a mistake in putting the limits(which is less probable) or i put the value of x instead of y to integrate. sorry for the mistake.

6. ## got the mistake!

Originally Posted by mybrohshi5
Looking at my original graph again of y=20-x^2 and y=x^2-12 i notice that there is more area above the x axis than below so it doesnt seem right to me that the area is 0. Just from looking at the graph it looks like the area should maybe be a positive number????
mistake in putting the limits:

there will be three integrals:

I1=integration(20,4)sqrt(y+12)dy
I2=integration(0,4)sqrt(20-y)dy
I3=integration(0,12)sqrt(20-y)dy

area=I1+I2+I3.

7. I don't see anything wrong with just doing
$\int_{-4}^{4}[(20-x^2)-(x^2-12)]dx$
as you say in your first post.

$\int_{-4}^4 (32- 2x^2) dx= \left[32x- 4/3 x^3\right]_{-4}^4$
which equals 85 and 2/3, not 170 and 2/3.

8. I believe you made a mistake on your integral. I think it should be

$
\int_{-4}^4 (32- 2x^2) dx= \left[32x- (2/3) x^3\right]_{-4}^4
$

which equals 170 and 2/3

Thank you for the help

9. Yes, of course!