Question

A small disposable drinking glass (turned sideways) can be modeled by rotating the curve
y=2*sqrt(x)
about the x-axis on the interval [1, 4]. Mickey needs to know the surface area of the outside of the cup.
He is able to compute the surface area of the base of the cup, but needs you and your calculus skills to
compute the surface area of the sides of the cup. Find this surface area, showing the integral you set
up and all symbolic work needed to evaluate the integral

Answer 1

Hello! Do you know the formula for this one? I had to look it up :P

Answer 2

@theEric y=\[y=2(\sqrt{x})\]

Answer 3

\[y=2\sqrt{x}\]*

Answer 4

Yep! That's the function that you will rotate about the \(x\) axis.

But, do you know what to do with it?

Answer 5

@theEric not at all man. haha. I know to integrate it but do rotations require taking the derivative of the function +1 all being under the sqrt and multiplied by 2pix for example, \[\int\limits_{1}^{4}2pix \sqrt{1+(f^\prime(x)})\]???? or am I wrong?

Answer 6

Sorry, I had to do something for a moment. You're very close!
\(\int\limits_{1}^{4}2\pi \color{blue}{f(x)} \sqrt{1+(f^\prime(x))\color{blue}{^2}\ \ }dx\\
\qquad\uparrow\qquad\qquad\qquad~\uparrow\)

Answer 7

And I get that from here:
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

Answer 8

So! Are you ready to tackle this?

Answer 9

I recommend finding \(f'(x)\) and then substituting it and \(f(x)\) in first.

Answer 10

Thank you so much! @theEric Ill let you know what I come up with!

Answer 11

Okay! And you're welcome!

Answer 12

i came out with 69.9688? is that correct?

Answer 13

Yep, that's what I got! Congrats!!