Question

My answer for this question was 4*Pi*(x^2+2)...
Find the area of the surface obtained by rotating the curve about the y-axis. y=2x^2+4, 0
10 years ago

Answer 1

You should get a single number and not a variable expression

Answer 2

I should?

Answer 3

yes, there shouldn't be an x in the final answer

Answer 4

Oh right... I'll check my working out again and will be right back..

Answer 5

\[\frac{ 1 }{ 128 }\pi(644 \sqrt{17}+127arcsinh(4))\]
That was what I got, was that what you obtained as well?

Answer 6

did you integrate this?
\[\Large \int_{0}^{1}2\pi x \sqrt{1+(4x)^2}dx\]

Answer 7

Not quite, I integrated this:
\[\int\limits_{0}^{1}2\pi(4+2x ^{2})\sqrt{1+16x^2}dx\]

Answer 8

Here's how I did it

\[\Large \int_{0}^{1}2\pi x \sqrt{1+(4x)^2}dx\]

\[\Large \int_{0}^{1}2\pi x \sqrt{1+16x^2}dx\]

Let u = 1+16x^2
du/dx = 32x ---> x*dx = du/32
if x = 0, then u = 1+16x^2 = 1+16(0)^2 = 1
if x = 1, then u = 1+16x^2 = 1+16(1)^2 = 17

\[\Large \int_{0}^{1}2\pi x \sqrt{1+16x^2}dx\]

\[\Large \int_{0}^{1}2\pi\sqrt{1+16x^2}*x*dx\]

\[\Large 2\pi\int_{1}^{17}\sqrt{u}*\frac{du}{32}\]

\[\Large 2\pi*\frac{1}{32}\int_{1}^{17}\sqrt{u}du\]

\[\Large \frac{\pi}{16}\int_{1}^{17}\sqrt{u}du\]

\[\Large \frac{\pi}{16}*\frac{2}{3}u^{{}^{3/2}}+C {\Huge]}_{1}^{17}\]

Plug in u = 17 and u = 1, then subtract. Do a bit of simplification to get

\[\Large \frac{\pi(17\sqrt{17}-1)}{24}\]

Answer 9

Notice how \(\Large \frac{\pi(17\sqrt{17}-1)}{24}\) is a single number and not a variable expression.

Answer 10

That's what I got if I were to integrate yours as well, right now I'm wondering why you got x where I got 4+2x^2

Answer 11

I'm basing things off this page

http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

specifically the stuff in the blue box

Answer 12

My friends are being helpful this time and reckon you're right, so thanks for your help! Much appreciated!! :)

Answer 13

you're welcome