# Volume of revolution

#### David S.

##### New member
Can someone please help me with this question

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#### Dr.Peterson

##### Elite Member
Sure.

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#### HallsofIvy

##### Elite Member
Imagine dividing the solid of rotation into many thin disks perpendicular to the x-axis. Each will have radius 2x+ 1 so what will be its area? The thickness will be "dx" so the volume is that area times dx. Integrate from 1 to 3.

(You don't really need Calculus. Imagine extending the upper and lower lines until they intersect. The top line is y= 2x+ 1 and the lower is y= -(2x+ 1) so they intersect when 2x+ 1= -2x- 1 or 4x= -2. They intersect when x= -1/2. That is a cone with base radius 2(3)+ 1= 7 and height 3+ 1/2= 3.5. What is the volume of that cone? [The volume of a cone with base radius r and height h is given by $$\displaystyle \frac{\pi}{3}r^2h$$.] Now you need to remove the part of that cone from x= -1/2 to 1. That is a cone with base radius 2(1)+ 1= 1= 3 and height 1.5. Calculate the volume of that smaller cone and subtract.)

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