Question

The given curve is rotated about the y-axis. Find the area of the resulting surface.
\[y=1-x^2\]
\[0\le x\le1\]
\[\frac{dy}{dx}=\frac{1}{2\sqrt{1-y}}\]
\[S=2\pi \int_{0}^{1} \sqrt{1-y}\sqrt{1+(\frac{1}{2\sqrt{1-y}})^2}dy\]
\[S=2\pi \int_{0}^{1} \sqrt{(1-y)+\frac12}dy\]
\[S=2\pi\left[\frac{-2}{3}(\frac12-y)^{\frac32}\right]_0^1\]
What did I do wrong?

Answer 1

would I just have
\[S=2\pi \int_{0}^{1} x\sqrt{1+(\frac{1}{2\sqrt{1-y}})^2}dx\]
?

Answer 2

sorry its been a long time since ive done this

Answer 3

what is the surface area equation

Answer 4

isn't it like f(x) * S where s is your arc length

Answer 5

\[S=\int 2\pi x ds\]

Answer 6

where \[ds=\sqrt{1+(\frac{dx}{dy})^2}dx\]

Answer 7

yeah that's the arc length

Answer 8

why is it just x instead of the equation x=......... \[S=2\pi \int_{0}^{1} x\sqrt{1+(\frac{1}{2\sqrt{1-y}})^2}dx\]

Answer 9

why didn't we sub anything for the x withing the integral?

Answer 10

I cannot remember thre is a reason behind it i just can't remember ... i believe it has to do with the radius

Answer 11

x is the length to the curve i think

Answer 12

like x= the length from the middle to the arc and ds is the arc length

Answer 13

before you put your dx/dy in ... why dont you square it

Answer 14

i think this might be where you aren't seeing something

Answer 15

the way that the book solved it is by leaving x as just plain and simple x....and then they just did the integral...

Answer 16

they did by dx?

Answer 17

yes....confusing :(

Answer 18

well for one if you change to y you have to change everything

Answer 19

that is probably why

Answer 20

so when I rotate about the y-axis it's still wrt y?

Answer 21

or the other way around? LOL I don't know anymore...

Answer 22

you can still write it in dx , it's not necessary i don't believe... if you switch to y, you have to replace x with a f(y)

Answer 23

and replace the limits by using the f(x)

Answer 24

you'll get the same answer so for your problem if you did that find (dx/dy)^2