# Thread: Area between two curves

1. ## Area between two curves

We just finished up area between two curves and volume of a solid. I've been having some difficulties with it. This was a homework problem two nights ago, and I thought I understood it (even after doing it in class), but I don't.

The question tells you to find the area between:

$4x+y^2=12$ and $x=y$

So I solve for x.

$4x=-y^2+12$

$x=\frac{-y^2+12}{4}$

Now...in class, the interval of the area was shown as -6 to 2 on the y axis. How is that determined? I can get -6 and 2 for values of intersection of x, but not y. What am I missing here?

The answer was eventually shown as (before computation):

$A= \int^{2} [(y) - (\frac{-y^2+12}{4})]dy$

I forgot how to put the lower part of the integral in there, sorry.

2. Originally Posted by Marconis
We just finished up area between two curves and volume of a solid. I've been having some difficulties with it. This was a homework problem two nights ago, and I thought I understood it (even after doing it in class), but I don't.

The question tells you to find the area between:

$4x+y^2=12$ and $x=y$

So I solve for x.

$4x=-y^2+12$

$x=\frac{-y^2+12}{4}$

Now...in class, the interval of the area was shown as -6 to 2 on the y axis. How is that determined? I can get -6 and 2 for values of intersection of x, but not y. What am I missing here?

The answer was eventually shown as (before computation):

$A= \int^{2} [(y) - (\frac{-y^2+12}{4})]dy$

I forgot how to put the lower part of the integral in there, sorry.
$4x + y^2 = 12$

and
$y = x$

So put y in for the x value:
$4y + y^2 = 12$

Solving for y gives you -6 and 2 as I'm sure you know. Now, y = x, so the x values at the points of intersection are the same as the y values. So the points of intersection are (-6., -6) and (2, 2).

-Dan

3. Originally Posted by Marconis

Now...in class, the interval of the area was shown as -6 to 2 on the y axis. How is that determined? I can get -6 and 2 for values of intersection of x, but not y. What am I missing here?
Take $\displaystyle y = \frac{12-y^2}{4}$

...

$\displaystyle (y-2)(y+6)=0$

4. Ah, okay. I was unsure if you could do that. Lol. *Sigh* math.

Thanks!

5. Sorry, one more thing...

I can't seem to recognize this graphically...any assistance on this?

6. Sketch $y=x$ and $y=\sqrt{12-4x}$ on the same axis, you are looking for the area in between.

7. Got it, thank you.

8. NEVERMIND, figured it out.