1. Area

Find c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.
c =?

2. Originally Posted by qbkr21
Find c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.
c =?
you are integrating with respect to x, so you cannot integrate the c terms like that. you would just attach x to them. by the way, when i did it your way i got 0 = 230, which means you probably made another mistake somewhere

3. Re:

Could you trying working it out? or at least some of it... so that I can get a feel for what you mean...

Sure appreciate it....

-qbkr21

4. Originally Posted by qbkr21
Could you trying working it out? or at least some of it... so that I can get a feel for what you mean...

Sure appreciate it....

-qbkr21
$\displaystyle \int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx$

$\displaystyle =2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx$

$\displaystyle = 4 \int_{0}^{c} \left( c^2 - x^2 \right)dx$

$\displaystyle = 4 \left[ c^2 x - \frac {1}{3}x^3 \right]_{0}^{c}$

$\displaystyle = \frac {8}{3}c^3$

check my computation, i was in a rush. i have to leave for a little while

did you see the difference with how i integrated c? i treated it as a constant--which it is

5. Re:

Yes you were right on the money as always!!

c = 4.41828

Thanks so much!!!

-qbkr21

6. find c

Originally Posted by Jhevon
$\displaystyle \int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx$

$\displaystyle =2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx$

$\displaystyle = 4 \int_{0}^{c} \left( c^2 - x^2 \right)dx$

$\displaystyle = 4 \left[ c^2 x - \frac {1}{3}x^3 \right]_{0}^{c}$

$\displaystyle = \frac {8}{3}c^3$

check my computation, i was in a rush. i have to leave for a little while

did you see the difference with how i integrated c? i treated it as a constant--which it is
I think you are right

7. Originally Posted by curvature
I think you are right
i know, qbkr21 already confirmed it. i believe he has the answer to the question in the back of his text. i hope he gets why i could have integrated between 0 and c though and then multiply by two, furthermore, i hope he gets why i would want to do that in the first place

8. Originally Posted by Jhevon
$\displaystyle \int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx$

$\displaystyle =2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx$
hi
what did you do to the 2 outside to make the lower integral -c 0?
can it be any constants?

hi
what did you do to the 2 outside to make the lower integral -c 0?
can it be any constants?
Note that your integrand is an even function: f(-x) = f(x). In this case
$\displaystyle \int_{-c}^c dx \, f(x) = \int_{-c}^0 dx \, f(x) + \int_0^c dx \, f(x)$

Use the substitution y = -x in the first integral:

$\displaystyle = \int_{c}^0 (-dy) \, f(-y) + \int_0^c dx \, f(x)$

$\displaystyle = -\left ( -\int_0^c dy \, f(-y) \right ) + \int_0^c dx \, f(x)$

But f(-y) = f(y) when f is an even function:

$\displaystyle = \int_0^c dy \, f(y) + \int_0^c dx \, f(x)$

and now just replace the dummy variable y in the first integration with x:

$\displaystyle = \int_0^c dx \, f(x) + \int_0^c dx \, f(x)$

$\displaystyle = 2 \int_0^c dx \, f(x)$

-Dan