Question

Help with a surface area question:

I'm trying to get the surface area of y=x^3/4 + 1/3x, from 1/2 < x < 1 about the x-axis, but I'm having trouble.

I'll post the integral I set up:

Answer 1

\[S = 2\pi \int\limits_{.5}^{1}\frac{ x^3 }{ 4 }+\frac{ 1 }{ 3x }*(\frac{ 3x^2 }{ 4 }+\frac{ 1 }{ 3x^2 })\]

Answer 2

Are you sure it's not x^(3/4)?

Answer 3

I FOIL'd that out so I get\[2\pi \int\limits_{.5}^{1} \frac{ 3x^5 }{ 16 }+\frac{ x^2 }{ 8 }+\frac{ x^2 }{ 24 }+\frac{ 1 }{ 9x^2 }\]

@madscientist101 Yeah, I'm sure. I just messed up and didn't use the right notation in my original post.

The original function with the correct notation is \[y= \frac{ x^3 }{ 4 }+\frac{ 1 }{ 3x }\]

Answer 4

hint: 1+(y')^2 is a perfect square

Answer 5

Are you sure you're not overcomplicating it? I did this math a long time ago... but I was able to find the integral of y.

Answer 6

\[\int\limits_{\frac{1}{2}}^{1}2 \pi y \sqrt{1+(y')^2} dx\]

Answer 7

and you'll take that plug in 1 and then .5 and then subtract the value of the expression with x=.5 from x=1...? Yes, no? I'm lost.

Answer 8

oh I see you meant \[S = 2\pi \int\limits\limits_{.5}^{1}(\frac{ x^3 }{ 4 }+\frac{ 1 }{ 3x })*(\frac{ 3x^2 }{ 4 }+\frac{ 1 }{ 3x^2 }) \]

Answer 9

@freckles Yeah I skipped the step where did 1+(dy/dx)^2 as a perfect square. My second post is the integral of 2pi * y * the square root of the perfect square. I think I've got that much right at least.


@madscientist101 No, there's a surface area forumula that you've got to use. @freckles posted it.

Answer 10

Ok well I just used up my last try on another guess. Turns out I was doing it right the whole time, and the online submission thing is just picky about the fractional answer it accepts. It was a crappy question anyway. Their example problem uses nice clean numbers that they know will make easy workable fractions, and then they give the students the ones that end up as 1981/6144.

Thanks for helping out guys.