Area of a Surface of Revolution Problem

[thatguy47]

Here's the problem:

Here's the solution to the problem:

I'm lost at where you get 2(pi)/36 in the solution. What is the reasoning for doing that? Can someone please explain it to me?

 

[galactus]

I agree with your solution.

$\displaystyle 2{\pi}\int_{a}^{b}g(y)\sqrt{1+[g'(y)]^{2}}dy$

$\displaystyle 2{\pi}\int_{1}^{2}y^{3}\sqrt{1+9y^{4}}dy$

Which gives 199.48, not Pi/18. Pi/18 is only equal to about .17. Surely the surface area is greater than that.

We could even set it up like so in terms of x and revolve about the y axis and get the same. I think there is a typo in the book.

$\displaystyle \frac{2\pi}{3}\int_{1}^{8}x\sqrt{x^{\frac{-4}{3}}+9} \;\ dx=199.48$

 

[galactus]

I'm sorry, I misunderstood. It would appear they multiplied the top and bottom by 36 in order to facilitate the algebra.

To inetgrate $\displaystyle \sqrt{1+9y^{4}}$, let $\displaystyle u=3y^{2}, \;\ du=6ydy$

Then, $\displaystyle \frac{u}{3}=y^{2}, \;\ \frac{du}{6}=ydy$

Put those in place of the $\displaystyle y^{3}dy$

 

[thatguy47]

I'm sorry, I misunderstood. It would appear they multiplied the top and bottom by 36 in order to facilitate the algebra.

To inetgrate $\displaystyle \sqrt{1+9y^{4}}$, let $\displaystyle u=3y^{2}, \;\ du=6ydy$

Then, $\displaystyle \frac{u}{3}=y^{2}, \;\ \frac{du}{6}=ydy$

Put those in place of the $\displaystyle y^{3}dy$

That makes sense. It confused me because they didn't plug in y=1 to y=2 into the u-substitution to get u=10 to u=145 which they plugged in on the last step. Thanks for the help.

 

[Deleted member 4993]

I think the actual substitution is:

$\displaystyle u \, = \, 1 \, + \, 9y^4$

$\displaystyle du \, = 36y^3 du$

They did not do the substitution explicitly - skipped few steps.