# Volumes of Revolution around the x-axis

• Feb 24th 2008, 09:21 AM
haku
Volumes of Revolution around the x-axis
The question is:
Find the volume generated when the finite region bounded by the given curves is rotated fully about the x-axis.

y^4=16x, x=1, x=4, y=0

If y^4=16x then y=2x - This is my curve.
Are the x values the limits?

Using the formula I can't get the correct answer. Can anyone help me out?

Pi * Integral(limits 4-1) of 2x
• Feb 24th 2008, 10:22 AM
earboth
Quote:

Originally Posted by haku
The question is:
Find the volume generated when the finite region bounded by the given curves is rotated fully about the x-axis.

y^4=16x, x=1, x=4, y=0

...

From

$\displaystyle y^4 = 16x$ you get $\displaystyle | y | = 2 \sqrt[4]{x}$

The volume is calculated by:

$\displaystyle V = \pi \int \left(f(x)\right)^2 dx$ That means here:

$\displaystyle V=\int_1^4\left(2 \cdot x^{\frac14} \right)^2 dx$

$\displaystyle V = \pi \left[4 \cdot \frac23 \cdot x^{\frac32} \right]_1^4$

I'll leave the rest for you.
• Feb 24th 2008, 10:23 AM
galactus
Washers:

$\displaystyle 4{\pi}\int_{1}^{4}\sqrt{x}dx$

shells:

$\displaystyle 4{\pi}\int_{2}^{2\sqrt{2}}\frac{y^{5}}{16}dy$
• Feb 24th 2008, 10:40 AM
haku
Okay so, [4.(2/3)x^(3/2)] Integrals 1 to 4

[64/3] - [8/3] = (56/3)*Pi

Is this correct?
• Feb 24th 2008, 11:27 AM
galactus
Yes, that's correct.
(Clapping)