Question

Did I do this right? Finding the surface area of curve y= x^2 (0,1) rotated about the y-axis.

Answer 1

\[SA= 2\pi \int\limits_{0}^{1}x(2x+1) dx\]

Answer 2

\[SA= 2\pi \int\limits_{0}^{1}2x^2 + x dx\]

Answer 3

\[SA= \frac{ 14\pi }{ 6 }\]

Answer 4

I simplified the arc length part

Answer 5

I'm not sure how you got the 2x+1 part. Can you show your work on that?

Hopefully you found that

y = x^2
dy/dx = 2x
[dy/dx]^2 = 4x^2
[dy/dx]^2+1 = 4x^2+1

agreed?

Answer 6

yes I agreed. I believe I made the mistake of assuming you could easily take the square root of that.

Answer 7

@jim_thompson5910

Answer 8

\[\Large \sqrt{4x^2+1} \ne \sqrt{4x^2} + \sqrt{1}\]

Answer 9

Ohh ok so plugging into the SA equation I have\[SA= 2\pi \int\limits_{0}^{1}x \sqrt{4x^2+1}dx\]

Answer 10

you have the correct integral now

Answer 11

ohh you can only do that if there weren't added right? like if they were seperate or a multiplication? what was inside the radical

Answer 12

yes
\[\Large \sqrt{x*y} = \sqrt{x}*\sqrt{y}\]

Answer 13

well, anyways moving on, I integrated using U substitution. I ended up with 5pi/4

Answer 14

let me check

Answer 15

http://www.wolframalpha.com/input/?i=2pi*int(x*sqrt(4x%5E2%2B1),x%3D0..1)

5pi/4 is not the correct answer

Answer 16

How do you integrate that then?

Answer 17

u = 4x^2+1
du = 8x*dx
x*dx = du/8

agreed so far?

Answer 18

yes, that's what I did

Answer 19

\[\Large 2\pi\int x\sqrt{4x^2+1}dx\]
\[\Large 2\pi\int (\sqrt{4x^2+1})*(x*dx)\]
\[\Large 2\pi\int \sqrt{u}*\frac{du}{8}\]
\[\Large 2\pi*\frac{1}{8}\int u^{1/2}du\]
\[\Large \frac{\pi}{4}\int u^{1/2}du\]
what comes next?

Answer 20

ohh I might of forgotten about the 3/2 power at the end :s

Answer 21

pi/4 * 2/3 * u^3/2

Answer 22

replace the u with 4x^2+1 to get

\[\Large \frac{\pi}{4}*\frac{2}{3}*(4x^2+1)^{3/2}+C\]

Answer 23

I get 5.85 now

Answer 24

close but not quite there

Answer 25

I think you forgot to plug in x = 0 and subtract

Answer 26

5.18

Answer 27

If
\[\Large g(x) = \Large \frac{\pi}{4}*\frac{2}{3}*(4x^2+1)^{3/2}+C\]
then what is g(1) equal to?

Answer 28

ohh algebra mistake, I get 5.33 now

Answer 29

5.33041350026898 which rounds to 5.33

so it looks good

Answer 30

Thanks :)

Answer 31

no problem

Answer 32

Now, if I want to find the SA of the same curve rotated about the x axis, the only thing that changes is the radius right?

Answer 33

\[SA= 2\pi \int\limits_{0}^{1}x^2\sqrt{4x^2+1}dx\]

Answer 34

You'll now use the formula
\[\Large SA = 2\pi\int_{a}^{b}y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\]

Answer 35

yes correct, evaluate what you just wrote

Answer 36

How would I integrate this? by parts?

Answer 37

Integration by parts

u = sqrt(4x^2+1)
du = 4x/sqrt(4x^2+1)


dv = x^2
v = (1/3)x^3


\[\Large \int u dv = u*v - \int v du\]

\[\Large \int x^2\sqrt{4x^2+1} = \sqrt{4x^2+1}*\frac{1}{3}x^3 - \int \frac{1}{3}x^3 \frac{4x}{\sqrt{4x^2+1}}dx\]

I'm getting stuck though

Answer 38

This person is using hyperbolic trig, which your teacher may have not gone over yet

http://mymathforum.com/calculus/48434-integral-x-2-sqrt-1-4x-2-a.html

Answer 39

so it's probably best to use a calculator to get the approx result

if you have a TI83, TI84, etc, you can use the function `FnInt` (which stands for Function Integration)

http://tibasicdev.wikidot.com/fnint

Answer 40

or you can use something like wolfram

http://www.wolframalpha.com/input/?i=2*pi*int(x%5E2*sqrt(x%5E2%2B1),x%3D0..1)