# Thread: Volume of solid revolution

1. ## Volume of solid revolution

Have a problems with the following question:

Consider the solid of revolution obtained when the region surrounded by x=3 ,x=6 ,y=0 and y= x+5 is rotated about the line y=2

Enter an expression for the cross-sectional area of this solid at a point

I can find the value..but how to get the expression?

I reckon the ans is suppose to be

((x+3)^2)*pi

but its wrong

2. Originally Posted by anthony162
Have a problems with the following question:

Consider the solid of revolution obtained when the region surrounded by x=3 ,x=6 ,y=0 and y= x+5 is rotated about the line y=2

Enter an expression for the cross-sectional area of this solid at a point

I can find the value..but how to get the expression?

I reckon the ans is suppose to be

((x+3)^2)*pi

but its wrong
1. I assume that you want to calculate the volume of the solid (?)

2. A trapezium rotates about the line y = 2 which yields a frustrum of a cone.

3. Use thin cylinders whose radius is $y-2=x+3$, whose thickness is dx. Use the formula to calculate the volume of a cylinder and add up all cylinders between x = 3 and x = 6.

4. $V=\int_3^6\left(\pi(x+3)^2\right)dx$

3. Nope..thats the second part of the question..which i have already answered..

4. Have you sketched it? I see a radius of x. I don't know where the 3 is coming from - probably formula-dazzle.