# Easy question but I forgot how to do it

#### 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!

#### skeeter

MHF Helper
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!
$$\displaystyle x^2-c^2 < c^2-x^2$$ for $$\displaystyle -c < x < c$$

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

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

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

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

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

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

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

s3a

#### 11rdc11

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!

Last edited:
s3a

#### drumist

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 $$\displaystyle -c<x<c$$, you of course know we set the two expressions equal:

$$\displaystyle x^2-c^2=c^2-x^2$$

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

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

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

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

$$\displaystyle \implies |x| = c$$

$$\displaystyle \implies x = \pm c$$

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

HallsofIvy and s3a