# Math Help - Another find the volume of a solid.Can I get myanswer checked please?

1. ## Another find the volume of a solid.Can I get myanswer checked please?

Find the volume of a solid generated by revolving the region bounded by the given lines adn curves about the x-axis.

$y=6x,y=6,y=0$

2. I get an answer of 12pi,is this correct?

Thanks

3. The question is vague: the region bounded by those lines is infinite and could be on either side of the line y = 6x.
The answer you gave is correct for the region bounded by the lines y = 6x, y = 0, and x = 1.

4. Maybe $y = 6x,{\text{ }}y = 6,{\text{ }}x = 0$ ??

5. Originally Posted by DeMath
Maybe $y = 6x,{\text{ }}y = 6,{\text{ }}x = 0$ ??
It was suppose to be x=0, not y=0.

6. So, you have ${y_1} = 6x,{\text{ }}{y_2} = 6,{\text{ }}x = 0,{\text{ }}{V_x} = ?$

Find the intersection point(s) of curves

${y_1} = {y_2} \Leftrightarrow 6x = 6 \Leftrightarrow x = 1.$

Then you have

${V_x} = \pi \int\limits_{x = a}^{x = b} {\left( {y_2^2 - y_1^2} \right)dx} = \pi \int\limits_0^1 {\left( {36 - 36{x^2}} \right)dx} = 36\pi \int\limits_0^1 {\left( {1 - {x^2}} \right)dx} =$

$= 36\pi \left. {\left( {x - \frac{{{x^3}}}{3}} \right)} \right|_0^1 = 36\pi \left( {1 - \frac{1}{3}} \right) = 24\pi {\text{ }}\left( {{\text{cubic units}}} \right).$

See this picture

7. Thanks for the picture. It really helps understanding it better!