# Math Help - Easy question but I forgot how to do it

1. ## Easy question but I forgot how to do it

Question:
Find such that the area of the region enclosed by the parabolas and is .

My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

Any help would be greatly appreciated!

2. Originally Posted by s3a
Question:
Find such that the area of the region enclosed by the parabolas and is .

My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

Any help would be greatly appreciated!
$x^2-c^2 < c^2-x^2$ for $-c < x < c$

$150 = \int_{-c}^c (c^2-x^2) - (x^2-c^2) \, dx$

$150 = 4\int_0^c c^2-x^2 \, dx$

$\frac{75}{2} = \left[c^2 x - \frac{x^3}{3}\right]_0^c$

$\frac{75}{2} = c^3 - \frac{c^3}{3}$

$\frac{75}{2} = \frac{2c^3}{3}$

$\frac{225}{4} = c^3$

$c = \sqrt[3]{\frac{225}{4}}$

3. Originally Posted by s3a
Question:
Find such that the area of the region enclosed by the parabolas and is .

My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

Any help would be greatly appreciated!

4. Originally Posted by s3a
Question:
Find such that the area of the region enclosed by the parabolas and is .

My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

Any help would be greatly appreciated!
If you are still wondering about why the limits of integration are $-c, you of course know we set the two expressions equal:

$x^2-c^2=c^2-x^2$

$\implies 2x^2 = 2c^2$

$\implies x^2 = c^2$

$\implies \sqrt{x^2} = \sqrt{c^2}$

Now, here is where most people make a mistake (or try to take shortcuts). What is the square root of $x^2$ ? It's $|x|$, NOT $x$. So our next step will be:

$\implies |x| = c$

$\implies x = \pm c$

By the way, the reason we can say that $\sqrt{c^2}=c$ is because the problem states that $c>0$.