V of solid rev..........

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February 26th 2010, 08:14 AM
wolfhound

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V of solid rev..........

Hi,
I have done the question on the picture attached
I don't believe it could be the correct answer, can someone help me please

Thanks
February 26th 2010, 08:28 AM
wolfhound

actually should my answer be 1114?
February 26th 2010, 09:19 AM
ione

Quote:

Originally Posted by wolfhound

Hi,
I have done the question on the picture attached
I don't believe it could be the correct answer, can someone help me please

Thanks

$(x^2+4x)^2=x^4+8x^3+16x^2$
February 26th 2010, 09:37 AM
wolfhound

Silly me,
Hows it looking now please?
still wrong I think
February 26th 2010, 01:49 PM
ione

Quote:

Originally Posted by wolfhound

Can someone help me with this please.....

That's correct to the nearest whole number.
February 26th 2010, 02:12 PM
wolfhound

but is it a rule to make it into a fraction and keep the pi(sign) or is it ok to do like I did
February 27th 2010, 07:15 AM
skeeter

$V = \pi \int_0^3 (x^2+4x)^2 \, dx<br />$

$V = \pi \int_0^3 x^4 + 8x^3 + 16x^2 \, dx<br />$

$V = \pi \left[\frac{x^5}{5} + 2x^4 + \frac{16x^3}{3}\right]_0^3<br />$

$V = \pi \left[\frac{3^5}{5} + 2 \cdot 3^4 + \frac{16 \cdot 3^3}{3}\right]<br />$

$V = \frac{1773\pi}{5} \approx 1114$
February 27th 2010, 07:15 AM
Archie Meade

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Quote:

Originally Posted by wolfhound

Is anyone there to help me with this annoying solid of revolution??????????
PLEASE!

Hi wolfhound.

It always helps to understand how the situation looks on a diagram.

The volume of revolution is the sum of the areas of a series of "wafer-thin" discs (an "infinite" sum of them, evaluated as their widths "deta-x" goes to zero).

The radius of the discs in this case is f(x) because the curve is being rotated around the x-axis.

the area of a disc is ${\pi}r^2={\pi}[f(x)]^2$

All of these disc areas are integrated from x=0 to x=3.
This gives the volume of revolution.

You need to get used to how to do this notationally,
as your written work needs to be redone.

$Disc\ area={\pi}\left(x^2+4x\right)^2={\pi}\left(x^4+8x^ 3+16x^2\right)$

Therefore, the volume of revolution is

$\int_{x=0}^{x=3}{\pi}\left(x^4+8x^3+16x^2\right)dx$

$={\pi}\left[\frac{x^5}{5}+\frac{8x^4}{4}+\frac{16x^3}{3}\right]$ from x=0 to 3

I've shown the disc from an angle (hence it looks like an ellipse!)
February 27th 2010, 10:09 AM
wolfhound

Thanks for the help, I must make more sense when I write it out..