anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • 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
  • TuringTest
yeah that's right so far, let me make sure I have the answer though...
TuringTest
  • 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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anonymous
  • 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
  • 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
  • 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
  • TuringTest
yeah it sure does... think about this: what is x^2 when x=0 ? what is 8-2x when x=0 ?
anonymous
  • 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
  • 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
  • anonymous
Well x^2 would be 0 and 8-2(0) would be 4
TuringTest
  • TuringTest
so 8-2x should be on the top, yes?
anonymous
  • anonymous
Oh so it should be (8-2x)-(x2) instead?
TuringTest
  • 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
  • TuringTest
...hence the negative answer
anonymous
  • anonymous
Okay let me try this and see if it is better
TuringTest
  • TuringTest
should be :D
anonymous
  • anonymous
dang I wish I knew all this like you ha would make this extensive homework easier
TuringTest
  • 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
  • anonymous
There is only one more question I had i doubt i can explain it here though. Let me try quick.
anonymous
  • 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
  • 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
  • TuringTest
I only see two shaded regions between these graphs: between -1,0 and 0,1
anonymous
  • anonymous
Our drawing is from -2 and 0, 0 to 3 ish? but some is above the x/3 line and some below
TuringTest
  • 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
  • anonymous
|dw:1328632999874:dw|
anonymous
  • anonymous
Not the best drawing but it wants the regions
TuringTest
  • TuringTest
click the link for a better image of the graph
TuringTest
  • TuringTest
does it tell you it wants the region between the graphs, or does it give you a specific interval to integrate?
anonymous
  • anonymous
The shaded region I drew. I think thats the region between graphs?
TuringTest
  • 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
  • anonymous
Well i think its bounded between the line and x=3
anonymous
  • anonymous
so he wants the area from the line at x=3 so it now has bounds i guess
TuringTest
  • TuringTest
then your intervals are\[[-1,0][0,1][1,3]\]
TuringTest
  • TuringTest
so you need to make sure that in each interval you have the proper graph on top, and proper graph on bottom
anonymous
  • anonymous
are u sure its not -2,0?
anonymous
  • anonymous
my book has the points (-2,2/3) as one point and (3,6) as another
TuringTest
  • TuringTest
if you want the region between the graphs then\[\frac{-2}{3}\neq\frac{(-2)^3}3\]so no
TuringTest
  • TuringTest
at least that's not a point of intersection
anonymous
  • anonymous
(-2,-2/3)
anonymous
  • anonymous
waitt
TuringTest
  • TuringTest
if you are told to integrate from x=-2 to x=3 then so be it, but those are not intersection points...
anonymous
  • anonymous
i feel stupid sorry. the line is x^3/3 -x
anonymous
  • anonymous
and x/3
TuringTest
  • TuringTest
ah... it happens
anonymous
  • anonymous
should make more sense
TuringTest
  • 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
  • anonymous
Well the 3rd shaded region the lines dont intercept
anonymous
  • anonymous
Its just from the point of (3,6) straight down to (3,1)
TuringTest
  • 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
  • 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
  • TuringTest
me too, got class you should be able to set up the integral now though, good luck!
anonymous
  • anonymous
Thanks, good luck
myname
  • myname
Is c a constant or a variable?

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