Question

find the surface area of the object formed by rotating the curve x = 1+y^2 for 1<=y <= 2 about the x-axis.

Answer 1

Arc length is given by:\[\large \int\limits ds\]The sum of all those little pieces of arc, \(\large ds\).
For surface area we take those little pieces of arc and spin them around.
So we're taking that arc length and multiplying it by the circumference that is formed by rotating.\[\large \int\limits (2\pi y)ds \qquad \text{or} \qquad \int\limits (2\pi x)ds\]

It looks like since we're spinning around the x-axis, we'll be using \(\large y\) as the radius of our object being spun around.\[\large 2\pi \int\limits y \;ds\]

Ok so what part are you stuck on? :)

Answer 2

One of the ways we can define ds is this way,
\[\large ds=\sqrt{1+\left(\frac{dx}{dy}\right)^2}\;dy\]

We'll want to use this particular method since it involves dy.
We want to integrate with respect to y.

Answer 3

So we need to find dx/dy.
Have you tried doing that yet?\[\large x=1+y^2\]\[\large x'=?\]

Answer 4

@zepdrix wait..if we are asked to find the area of that object, shouldn't we use \[\int\limits_{a}^{b} f(x)(\sqrt(1+（f'(x)))\]?

Answer 5

Hmm your formula looks a little off.
I'm not sure where that f(x) is coming from.
If we were integrating with respect to x it would be:\[\large \int\limits_a^b 2\pi x\sqrt{1+f'(x)^2}\;dx\]

But we're integrating with respect to y, so it'll look a little different.\[\large \int\limits\limits_a^b 2\pi y\sqrt{1+f'(y)^2}\;dx\]

Answer 6

\[\large \int\limits\limits\limits_a^b 2\pi y\sqrt{1+f'(y)^2}\;dy\]Woops dy at the end :O

Answer 7

oh right!...i remembered wrong...D:

So after sub in all the values i got \[2\Pi(\int\limits_{5}^{2} \sqrt{x-1} \sqrt{1+ (1/4(x-1))}\] am I right?

Answer 8

Oh you chose to change everything to x and integrate that way? Weird +_+ lol
Hmm yah looks correct so far! :)\[\large 2\pi\int\limits_2^5 \sqrt{x-1}\sqrt{1+\frac{1}{4(x-1)}}\;dx\]

Answer 9

thank you so much :D I guess I can do the rest on my on. :D

Answer 10

ok cool c: