## anonymous 4 years ago 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.

1. 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.

2. TuringTest

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

3. 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$=2\int_{0}^{c}2c^2-2x^2=2(2c^2x-\frac23x^3)|_{0}^{c}=2(2c^3-\frac23c^3)=72$

4. 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.

5. 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!

6. 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?

7. TuringTest

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

8. 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.

9. 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

10. anonymous

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

11. TuringTest

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

12. anonymous

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

13. TuringTest

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

14. TuringTest

...hence the negative answer

15. anonymous

Okay let me try this and see if it is better

16. TuringTest

should be :D

17. anonymous

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

18. 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

19. anonymous

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

20. 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

21. 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

22. TuringTest

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

23. anonymous

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

24. 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

25. anonymous

|dw:1328632999874:dw|

26. anonymous

Not the best drawing but it wants the regions

27. TuringTest

click the link for a better image of the graph

28. TuringTest

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

29. anonymous

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

30. 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

31. anonymous

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

32. anonymous

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

33. TuringTest

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

34. TuringTest

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

35. anonymous

are u sure its not -2,0?

36. anonymous

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

37. TuringTest

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

38. TuringTest

at least that's not a point of intersection

39. anonymous

(-2,-2/3)

40. anonymous

waitt

41. TuringTest

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

42. anonymous

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

43. anonymous

and x/3

44. TuringTest

ah... it happens

45. anonymous

should make more sense

46. 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 :/

47. anonymous

Well the 3rd shaded region the lines dont intercept

48. anonymous

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

49. 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

50. 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

51. TuringTest

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

52. anonymous

Thanks, good luck

53. myname

Is c a constant or a variable?