Question

Can anyone help me with a calculus surface area problem?

Answer 1

Here is the problem I am stuck on: http://imgur.com/RJtqe

Answer 2

area = \[\int\limits_{x,}^{x,,}y.dx\]
here inside the integral
[x^(1/3)+2]dx
on integrating we get
x^(4/3)*3/4+2x
then put the limit x, to x,,

Answer 3

It looks like the disk method works best here. We get volume elements by slicing parallel to the x-y plane to get horizontal circular cross sections of thickness dy, radius r, for now. That element has volume pi*r^2 dy. Let's get r:
Solve the original equation for x to get
x = (y-2)^3, and that is the radius of the disk. Then
\[V = \pi \int\limits_{2}^{3}(y - 2)^{6} dy.\]

Answer 4

It should be:

\[\text{Surface rotated about y-axis} = \int_a^b2\pi x\sqrt{1+(\frac{dx}{dy})^2} dy \]

Answer 5

Where:
\(y=\sqrt[3]{x} +2 \)\[\implies x = (y-2)^3\]\[\implies \frac{dx}{dy} = 3(y-2)^2\]

Answer 6

And your y values are going from 3 to 4.

Answer 7

Surface area. My glitch.

Answer 8

What I tried was:
since it is rotated about y axis, rewrite function in terms of y to get: h(y)=\[\left(y-2\right)^{3}\]
take that derivative to get: h'(y) = \[3\left( y-2 \right)^{2}\]
plug h(y) and h'(y) bounds etc. into the formula here: http://imgur.com/9IsNR
To get: \[2\pi \int\limits_{1}^{8}\left( y-2 \right)^{3}\sqrt{1+\left( 9(y-4 \right)^{4})}\]

Substitute: u = \[1 + 9(y-2)^{4}\]
du = \[36(y-2)^{3}\] so... du/36= \[(y-2)^{3}\]

plug substitute u and du/36 back into equation with new bounds (in terms of u) to (after summarizing) get: \[\pi/18\int\limits_{10}^{11665} du \sqrt{u}\]

after integration and then evaluation yields a massive number: 146589.3173, this cant be right

Answer 9

Looks like you did nearly everything correctly, but you used the wrong limits. You are integrating with respect to y, and your y values go from 3 to 4, not from 1 to 8.

Answer 10

Ah ha! that was it! Knowing when to use what limits / what variation of the function really is the hardest thing about these problems! thanks!

Answer 11

Yeah, getting everything right when you set them up is definitely the trick =)