Thread: volume of solid of revolution further integration

1. 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

2. Re: volume of solid of revolution further integration

Originally Posted by bigmansouf
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.

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 $\pi$?

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.

3. Re: volume of solid of revolution further integration

$\displaystyle V = \pi \int_0^3 y^2 \, dx$

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

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

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

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

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

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

$\displaystyle \bar{x} = \dfrac{1}{2}\bigg[\left(\frac{9}{2} - \frac{9}{4} \right)-(0) \bigg] = \dfrac{9}{8}$

4. Re: volume of solid of revolution further integration

Originally Posted by Walagaster
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

5. Re: volume of solid of revolution further integration

Originally Posted by bigmansouf
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).

6. Re: volume of solid of revolution further integration

Originally Posted by Walagaster
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