Volume of revolution
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Dr.Peterson
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Sure.
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HallsofIvy
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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.)
(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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