Question

So this question had me stumped on my test earlier.
Find the surface area:
y=sqrt(1-(x^2/4)) where 0 is less than or equal to x less than or equal to 2

Answer 1

\[y=\sqrt{1-x^2/4}\]

Answer 2

\[0\le x \le2\]

Answer 3

surface area?
is this a revolution?

Answer 4

And the formula is
\[\int\limits_{a}^{b}2\Pi yds\]

Answer 5

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

Answer 6

so you are revolving around the x-axis?

Answer 7

Yesh, sorry for not answering

Answer 8

about the x-axis

Answer 9

Ok here's an idea...

Answer 10

Any ideas are much appreciated!

Answer 11

\[y=\sqrt{1-\frac{x^2}4}\implies y^2=1-\frac{x^2}4\]using implicit differentiation we get\[2yy'=-\frac x2\implies y'=-{x\over4y}\]\[2\pi\int yds=2\pi\int y\sqrt{1+{x^2\over16y^2}}dx=2\pi\int_{0}^{2}\sqrt{y^2+{x^2\over 16}}dx\]I'll continue, but let you digest in the meantime...

Answer 12

Sounds good, willing to wait

Answer 13

Wait, is not ok to leave it in terms of x since we are rotating about the x-axis?? and by "y" in the formula, it means whatever y=. Just to clear up some confusion.

Answer 14

but this is where the trick of implicit differentiation pays off: we will now sub in that expression for y into the formula above:\[2\pi\int_{0}^{2}\sqrt{1-{x^2\over4}+{x^2\over16}}dx=2\pi\int_{0}^{2}\sqrt{1-{3x^2\over16}}dx\]and now we have a trig sub... I hope

Answer 15

more specifically we subbed in for \(y^2\)

Answer 16

Haha oh ok. I was thinking I would need a trig substitution, but radical was not very pretty looking

Answer 17

yeah, but waiting to sub in for y made it a lot easier

Answer 18

cause my radical looked very similar but my trig substitution was not making any sense. This is what i had on my test, which I know is wrong.
\[\pi/2\int\limits_{0}^{2} \sqrt{-3x^2+16}\]

Answer 19

it is the same as mine, but with pi/2 for some reason I don't get...

Answer 20

Well I had 16 on bottom, so i split the radical up into to with numerator squared over the denominator squared. And the sqrt(16) is 4 so I took out the constant of 1/4 and that times 2pi=pi/2. But my trig sub. was not working out to nicely, so I figured I messed up somewhere.

Answer 21

oh yeah, my algebra was off
you are right about the pi/2
that said, we have the same thing\[\sqrt{-3x^2+16}=\sqrt{16-3x^3}=\frac14\sqrt{1-{3x^2\over16}}\]so try the sub\[\frac4{\sqrt3}\sin\theta=x\]

Answer 22

ohhhh that is where I messed up, all i did was
\[1/\sqrt{3}\] \[\sin \theta\]

Answer 23

well those are suppose to be together. haha

Answer 24

gotta get rid of that 16 ;)
I gotta go, glad I could help!

Answer 25

Yes, thank you very much!!!