# volume of solid of revolution further integration

• Mar 20th 2019, 10:00 AM
bigmansouf
volume of solid of revolution further integration
Question: An area in the first quadrant is bounded by the ellipse $\displaystyle 4x^2+9y^2=36$ and the axes. This area is rotated through four right angles about the x-axis. Find
(a) the volume of the solid generated
(b) the x-coordinate of the centre of gravity of this solid

My attempt:
a) $\displaystyle 4x^2+9y^2=36$

$\displaystyle \frac{x^2}{9}+ \frac{y^2}{4 }=1$

$\displaystyle y =\frac{\sqrt{4-\frac{4}{9}x^2}}{2}$
element of volume = $\displaystyle \pi y^2 \partial x$
v =$\displaystyle \int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x$

v =$\displaystyle 2$
b) $\displaystyle \bar{x}\int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x = \int_{0}^{3} \pi x \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x$

$\displaystyle \bar{x}\int_{0}^{3} \left ( \frac{{4-\frac{4}{9}x^2}}{4} \right ) \mathrm{d} x = \int_{0}^{3} x - \frac{x^3}{9} \mathrm{d} x$

$\displaystyle \bar{x} = \frac{9}{8}$

but the book gives the answer for a) $\displaystyle 8 \pi$ b) $\displaystyle \frac{4}{5}$

please can i have some help with this question
• Mar 20th 2019, 11:15 AM
Walagaster
Re: volume of solid of revolution further integration
Quote:

Question: An area in the first quadrant is bounded by the ellipse $\displaystyle 4x^2+9y^2=36$ and the axes. This area is rotated through four right angles about the x-axis. Find
(a) the volume of the solid generated
(b) the x-coordinate of the centre of gravity of this solid

My attempt:
a) $\displaystyle 4x^2+9y^2=36$

$\displaystyle \frac{x^2}{9}+ \frac{y^2}{4 }=1$

$\displaystyle y =\frac{\sqrt{4-\frac{4}{9}x^2}}{2}$

Check your algebra solving for $y$ here.

Quote:

element of volume = $\displaystyle \pi y^2 \partial x$
v =$\displaystyle \int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x$

v =$\displaystyle 2$
What happened to the $\pi$?

Quote:

b) $\displaystyle \bar{x}\int_{0}^{3} \pi \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x = \int_{0}^{3} \pi x \left ( \frac{\sqrt{4-\frac{4}{9}x^2}}{2} \right )^2 \mathrm{d} x$

$\displaystyle \bar{x}\int_{0}^{3} \left ( \frac{{4-\frac{4}{9}x^2}}{4} \right ) \mathrm{d} x = \int_{0}^{3} x - \frac{x^3}{9} \mathrm{d} x$

$\displaystyle \bar{x} = \frac{9}{8}$

but the book gives the answer for a) $\displaystyle 8 \pi$ b) $\displaystyle \frac{4}{5}$

please can i have some help with this question
Please don't use color and font commands throughout like that because it makes it very difficult to edit.
• Mar 20th 2019, 02:04 PM
Cervesa
Re: volume of solid of revolution further integration
$\displaystyle V = \pi \int_0^3 y^2 \, dx$

$\displaystyle V = \pi \int_0^3 4\left(1 - \frac{x^2}{9} \right) \, dx$

$\displaystyle V = 4\pi \bigg[x - \frac{x^3}{27} \bigg]_0^3$

$\displaystyle V = 4\pi \bigg[\left(3 - \frac{3^3}{27}\right) - (0) \bigg] = 8\pi$

$\displaystyle \bar{x} = \dfrac{\pi \int_0^3 x \cdot y^2 \, dx}{\pi \int_0^3 y^2 \, dx} = \dfrac{4\pi \int_0^3 x - \frac{x^3}{9} \, dx}{8\pi}$

$\displaystyle \bar{x} = \dfrac{1}{2} \int_0^3 x - \frac{x^3}{9} \, dx$

$\displaystyle \bar{x} = \dfrac{1}{2}\bigg[\frac{x^2}{2} - \frac{x^4}{36} \bigg]_0^3$

$\displaystyle \bar{x} = \dfrac{1}{2}\bigg[\left(\frac{9}{2} - \frac{9}{4} \right)-(0) \bigg] = \dfrac{9}{8}$
• Mar 20th 2019, 04:07 PM
bigmansouf
Re: volume of solid of revolution further integration
Quote:

Check your algebra solving for $y$ here.

What happened to the $\pi$?

Please don't use color and font commands throughout like that because it makes it very difficult to edit.

sorry but cannot you explain what you mean by font commands and color
thank you
• Mar 20th 2019, 09:33 PM
Walagaster
Re: volume of solid of revolution further integration
Quote:

sorry but cannot you explain what you mean by font commands and color
thank you

Try "reply with quote" to your original post and I think you will see what I am talking about. (I think so, anyway).
• Mar 21st 2019, 02:06 AM
topsquark
Re: volume of solid of revolution further integration
Quote:

Try "reply with quote" to your original post and I think you will see what I am talking about. (I think so, anyway).

Wow.

-Dan