# area of the surface obtained by rotating the curve

• Apr 28th 2010, 09:00 PM
nhatie
area of the surface obtained by rotating the curve
4) Find the area of the surface obtained by rotating the curve from http://webwork.umemat.maine.edu/webw...c67fef88e1.png to http://webwork.umemat.maine.edu/webw...286ca42401.png about the http://webwork.umemat.maine.edu/webw...dd0b8b8e91.png-axis.

Any idea about the answer? I try to do it and my answer always turn out wrong. I don't know why:(
• Apr 28th 2010, 09:41 PM
Amer
Quote:

Originally Posted by nhatie
4) Find the area of the surface obtained by rotating the curve from http://webwork.umemat.maine.edu/webw...c67fef88e1.png to http://webwork.umemat.maine.edu/webw...286ca42401.png about the http://webwork.umemat.maine.edu/webw...dd0b8b8e91.png-axis.

Any idea about the answer? I try to do it and my answer always turn out wrong. I don't know why:(

$\displaystyle S = \int_{a}^{b} 2\pi f(x) \sqrt{1+(f'(x))^2} dx$ around x-axis

so we need to find dy/dx
$\displaystyle dy/dx = \frac{5}{2\sqrt{5x}}$

the surface area

$\displaystyle S = 2\pi \int_{0}^{4} \sqrt{5x} \sqrt{1+ \frac{25}{4(5x)}} dx$
• Apr 29th 2010, 05:43 AM
nhatie
Quote:

Originally Posted by Amer
$\displaystyle S = \int_{a}^{b} 2\pi f(x) \sqrt{1+(f'(x))^2} dx$ around x-axis

so we need to find dy/dx
$\displaystyle dy/dx = \frac{5}{2\sqrt{5x}}$

the surface area

$\displaystyle S = 2\pi \int_{0}^{4} \sqrt{5x} \sqrt{1+ \frac{25}{4(5x)}} dx$

I worked out to that part and got stuck there? How can you solve it?
• Apr 29th 2010, 06:58 AM
Amer
Quote:

Originally Posted by nhatie
I worked out to that part and got stuck there? How can you solve it?

I will not solve it

$\displaystyle S = 2\pi \int_0^4 \sqrt{5x + \frac{25}{4} } dx$

you should know how to integrate this
• Apr 29th 2010, 07:00 AM
nhatie
Quote:

Originally Posted by Amer
I will not solve it

$\displaystyle S = 2\pi \int_0^4 \sqrt{5x + \frac{25}{4} } dx$

you should know how to integrate this

i got : -2(5x +25/4)^-1/2 from 0 to 4 , is it rite??
• Apr 29th 2010, 07:10 AM
Amer
drive $\displaystyle -2(5x +25/4)^{-1/2}$ it should equal $\displaystyle \sqrt{5x + \frac{25}{4} }$ if it is true
• Apr 29th 2010, 07:16 AM
HallsofIvy
Quote:

Originally Posted by nhatie
i got : -2(5x +25/4)^-1/2 from 0 to 4 , is it rite??

Almost but not quite.

Let u= 5x+ 25/4. Then du= 5dx so that dx= (1/5)du and the integral is $\displaystyle (1/5)\int u^{1/2}du$. You don't have the "1/5" and the integral of $\displaystyle u^{1/2}$ is [tex](2/3)u^{3/2}[/itex]. It looks like you differentiated rather that integrating.
• Apr 29th 2010, 10:33 AM
nhatie
Quote:

Originally Posted by HallsofIvy
Almost but not quite.

Let u= 5x+ 25/4. Then du= 5dx so that dx= (1/5)du and the integral is $\displaystyle (1/5)\int u^{1/2}du$. You don't have the "1/5" and the integral of $\displaystyle u^{1/2}$ is [tex](2/3)u^{3/2}[/itex]. It looks like you differentiated rather that integrating.

ok i have it now and i have this
1/5 int 2/3u^(3/2) => 1/5(2/3(5x+25/4)^3/2) |from 0 to 4
• Apr 29th 2010, 10:39 AM
Amer
correct solution
• Apr 29th 2010, 10:41 AM
nhatie
Quote:

Originally Posted by Amer
correct solution

but when i put it in.. the answe is wrong? can you recaculate my work?
• Apr 29th 2010, 10:46 AM
Amer
Quote:

Originally Posted by nhatie
but when i put it in.. the answe is wrong? can you recaculate my work?

multiply it with $\displaystyle 2\pi$ (Wink)
• Apr 29th 2010, 10:51 AM
nhatie
Quote:

Originally Posted by Amer
multiply it with $\displaystyle 2\pi$ (Wink)

i have 99.5 but it is still wrong!!