Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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!

Answer 6

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?

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

...hence the negative answer

Answer 15

Okay let me try this and see if it is better

Answer 16

should be :D

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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