Question

Surface area problem with integration

Answer 1

The region between \[y=(x^3/3) , x=(3^(1/4)), x=15^(1/4)\] is revolved about the x-axis

Answer 2

Find the surfaced area.

Answer 3

I don't know how to do this, but I would start by reading this:
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

Answer 4

thanks

Answer 5

the region between\[y=\frac{x^3}3\]\[x=3^{1/4}\]and\[x=5^{1/4}\]revolved around x?

Answer 6

yes

Answer 7

No \[15^{1/4}\]

Answer 8

the region between\[y=\frac{x^3}3\]\[x=3^{1/4}\]and\[x=15^{1/4}\]

Answer 9

yes

Answer 10

first you need \[ds=\sqrt{1+(\frac{dy}{dx})^2}\]

Answer 11

That should be equal to \[\sqrt{1+x^4}\]

Answer 12

right, then this is going around the x-axis
so what are we multiplying this by?

Answer 13

\[x^3\]

Answer 14

\[y=\frac{x^3}3\]

Answer 15

and also by \(2\pi\)

Answer 16

right sorry. but I factored out 2pi and 1/3

Answer 17

so our integral is starting to shape up\[\int_a^b2\pi yds=\frac{2\pi}3\int x^3\sqrt{1+x^4}dx\]and the bounds are...?

Answer 18

exactly I got just that

Answer 19

and then i did a substitution and got new bounds

Answer 20

sorry, did you want to continue here?

Answer 21

yes please, I just want to make sure I am correct. My final answer is \[224\pi/27\]

Answer 22

let me try\[u=x^4+1\implies du=4x^3dx\]so the new bounds are\[u=4\to u=16\]\[\frac{\pi}6\int_4^{16}\sqrt udu=\frac\pi9u^{3/2}|_4^{16}=\frac\pi9(64-8)=\frac{56\pi}9\]so I get a somewhat different answer
I dunno, do you see a mistake?

Answer 23

how did you get pi/6?

Answer 24

Im going to work the problem over. I probably made a mistake somewhere.

Answer 25

\[\frac14du=x^3dx\]\[\frac{2\pi}3\int_4^{16}(\frac14du)\sqrt udu=\frac\pi6\int_4^{16}\sqrt udu\]

Answer 26

Yea I found my mistake sorry

Answer 27

right-on :)