Find the postive value of c such that the area of the the region enclosed by the parabolas y = x^2 -c^2 and y = c^2 -x^2 is 72.

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Find the postive value of c such that the area of the the region enclosed by the parabolas y = x^2 -c^2 and y = c^2 -x^2 is 72.

Calculus1
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I thought I had to show c^2=x^2 so x=+-c so I evaluate from -c to c. But the equation is throwing me off if I was even right.
yeah that's right so far, let me make sure I have the answer though...
ok, we got the top graph is c^2-x^2, and the bottom is x^2-c^2, so our integral is\[\int_{-c}^{c}c^2-x^2-(x^2-c^2)dx=\int_{-c}^{c}2c^2-2x^2dx\]the integrand is even so we can simplify this a bit\[=2\int_{0}^{c}2c^2-2x^2=2(2c^2x-\frac23x^3)|_{0}^{c}=2(2c^3-\frac23c^3)=72\]

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ohh I was doing it the other way. I put f(x) to be (x^2-c^2) and g(x) to be the other one and was getting a werid answer.
yeah, it's a really good idea to sketch these thing out when you're not 100% on them. many people neglect that, and it costs them. welcome!
Speaking of sketching, can you check what I may have done wrong here. It says y=x^2 and y=8-2x find area between x-axis and 2 curves. So I did \[\int\limits_{-4}^{2}x^2-(8-2x)\] it gives me a negative area?
yeah it sure does... think about this: what is x^2 when x=0 ? what is 8-2x when x=0 ?
So wait did I do it wrong? I thought I should find where they intercept and then take the intergral. Iver never done one like this with the "between x-axis" part so not sure.
yes, but just answer my question let f(x)=x^2 let g(x)=8-2x what is f(0) ? what is g(0) ? picking a point in your interval of integration to check will tell you which graph should be on top and which should be on the bottom
Well x^2 would be 0 and 8-2(0) would be 4
so 8-2x should be on the top, yes?
Oh so it should be (8-2x)-(x2) instead?
yep 8-2x>x^2 on the interval you are talking about http://www.wolframalpha.com/input/?i=plot+x%5E2%2C8-2x
...hence the negative answer
Okay let me try this and see if it is better
should be :D
dang I wish I knew all this like you ha would make this extensive homework easier
Just practice is all. What I study is still hard for me 'cause it's new. By the time you learn it it's on to the next thing :P
There is only one more question I had i doubt i can explain it here though. Let me try quick.
When you are finding the area of 2 lines with 3 shaded regions, how do you find the middle one when points arent given? the lines are y=x/3 and y=x^3/3
I can get the first region because i just evaluate from -2 to 0. The rest never seen before since some is above the x/3 line and some before
I only see two shaded regions between these graphs: between -1,0 and 0,1
Our drawing is from -2 and 0, 0 to 3 ish? but some is above the x/3 line and some below
\[\frac x3=\frac{x^3}3\to x=x^3\to x(1-x^2)=0\to x=\left\{ -1,0,1 \right\}\]are the intervals here's a visual http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3
|dw:1328632999874:dw|
Not the best drawing but it wants the regions
click the link for a better image of the graph
does it tell you it wants the region between the graphs, or does it give you a specific interval to integrate?
The shaded region I drew. I think thats the region between graphs?
but the last region you drew on the right is unbounded: it goes to infinity again click the link I posted, there are only 2 regions enclosed between the graphs: http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3
Well i think its bounded between the line and x=3
so he wants the area from the line at x=3 so it now has bounds i guess
then your intervals are\[[-1,0][0,1][1,3]\]
so you need to make sure that in each interval you have the proper graph on top, and proper graph on bottom
are u sure its not -2,0?
my book has the points (-2,2/3) as one point and (3,6) as another
if you want the region between the graphs then\[\frac{-2}{3}\neq\frac{(-2)^3}3\]so no
at least that's not a point of intersection
(-2,-2/3)
waitt
if you are told to integrate from x=-2 to x=3 then so be it, but those are not intersection points...
i feel stupid sorry. the line is x^3/3 -x
and x/3
ah... it happens
should make more sense
\[\frac{x^3}3-x=\frac x3\to\frac{x^3}3=\frac{4x}{3}\to x(x^2-4)=0\]\[x=\left\{ -2,0,2 \right\}\]so it's still not the right interval :/
Well the 3rd shaded region the lines dont intercept
Its just from the point of (3,6) straight down to (3,1)
ok, so we have from\[[-2,0][0,2][2,3]\]look at the graph http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3-x on the first interval x^3/3-x is on top on the second, x/3 is on top on the third, x^3/3 is on top again
I have to go eat lunch, thanks for the help though. I will run it past my professor and maybe will get help before its due. Thanks
me too, got class you should be able to set up the integral now though, good luck!
Thanks, good luck
Is c a constant or a variable?

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