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.

- anonymous

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- anonymous

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.

- TuringTest

yeah that's right so far, let me make sure I have the answer though...

- TuringTest

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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## More answers

- anonymous

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.

- TuringTest

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!

- anonymous

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?

- TuringTest

yeah it sure does...
think about this: what is x^2 when x=0 ?
what is 8-2x when x=0 ?

- anonymous

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.

- TuringTest

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

- anonymous

Well x^2 would be 0 and 8-2(0) would be 4

- TuringTest

so 8-2x should be on the top, yes?

- anonymous

Oh so it should be (8-2x)-(x2) instead?

- TuringTest

yep
8-2x>x^2 on the interval you are talking about
http://www.wolframalpha.com/input/?i=plot+x%5E2%2C8-2x

- TuringTest

...hence the negative answer

- anonymous

Okay let me try this and see if it is better

- TuringTest

should be :D

- anonymous

dang I wish I knew all this like you ha would make this extensive homework easier

- TuringTest

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

- anonymous

There is only one more question I had i doubt i can explain it here though. Let me try quick.

- anonymous

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

- anonymous

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

- TuringTest

I only see two shaded regions between these graphs:
between -1,0 and 0,1

- anonymous

Our drawing is from -2 and 0, 0 to 3 ish? but some is above the x/3 line and some below

- TuringTest

\[\frac x3=\frac{x^3}3\to x=x^3\to x(1-x^2)=0\to x=\left\{ -1,0,1 \right\}\]are the intervals
here's a visual
http://www.wolframalpha.com/input/?i=plot+x%2F3%2Cx%5E3%2F3

- anonymous

|dw:1328632999874:dw|

- anonymous

Not the best drawing but it wants the regions

- TuringTest

click the link for a better image of the graph

- TuringTest

does it tell you it wants the region between the graphs, or does it give you a specific interval to integrate?

- anonymous

The shaded region I drew. I think thats the region between graphs?

- TuringTest

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

- anonymous

Well i think its bounded between the line and x=3

- anonymous

so he wants the area from the line at x=3 so it now has bounds i guess

- TuringTest

then your intervals are\[[-1,0][0,1][1,3]\]

- TuringTest

so you need to make sure that in each interval you have the proper graph on top, and proper graph on bottom

- anonymous

are u sure its not -2,0?

- anonymous

my book has the points (-2,2/3) as one point and (3,6) as another

- TuringTest

if you want the region between the graphs then\[\frac{-2}{3}\neq\frac{(-2)^3}3\]so no

- TuringTest

at least that's not a point of intersection

- anonymous

(-2,-2/3)

- anonymous

waitt

- TuringTest

if you are told to integrate from x=-2 to x=3 then so be it, but those are not intersection points...

- anonymous

i feel stupid sorry. the line is x^3/3 -x

- anonymous

and x/3

- TuringTest

ah...
it happens

- anonymous

should make more sense

- TuringTest

\[\frac{x^3}3-x=\frac x3\to\frac{x^3}3=\frac{4x}{3}\to x(x^2-4)=0\]\[x=\left\{ -2,0,2 \right\}\]so it's still not the right interval :/

- anonymous

Well the 3rd shaded region the lines dont intercept

- anonymous

Its just from the point of (3,6) straight down to (3,1)

- TuringTest

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

- anonymous

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

- TuringTest

me too, got class
you should be able to set up the integral now though, good luck!

- anonymous

Thanks, good luck

- myname

Is c a constant or a variable?

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