Need help with finding a volume of revolution with Cylindrical Shells method

y=2-0.5x, y=0, x=1, x=2, rotate about the x axis.

I know that it will be easier to solve it with the regular method, but I need it with cylindrical shell.

I cannot find the radius and the height.

The final answer is 19pi/12

Thanx

Re: Need help with finding a volume of revolution with Cylindrical Shells method

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Originally Posted by

**shaharg** y=2-0.5x, y=0, x=1, x=2, rotate about the x axis.

I know that it will be easier to solve it with the regular method, but I need it with cylindrical shell.

I cannot find the radius and the height.

If you must use the shell method, you will need to set up two integrals: one for $\displaystyle 0\leq y\leq 1,$ in which the height of each shell is simply 1, and then another for $\displaystyle \textstyle1\leq y\leq\frac32,$ in which the height of each shell is $\displaystyle (4-2y)-1.$

Re: Need help with finding a volume of revolution with Cylindrical Shells method

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Originally Posted by

**Reckoner** If you must use the shell method, you will need to set up two integrals: one for $\displaystyle 0\leq y\leq 1,$ in which the height of each shell is simply 1, and then another for $\displaystyle \textstyle1\leq y\leq\frac32,$ in which the height of each shell is $\displaystyle (4-2y)-1.$

So, what is the radius and what is the height?

Re: Need help with finding a volume of revolution with Cylindrical Shells method

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Originally Posted by

**shaharg** So, what is the radius and what is the height?

I suggest drawing a picture. The average radius of each shell is the distance from the axis of rotation to each given $\displaystyle y$-value. Since the axis of rotation is simply the $\displaystyle x$-axis, this distance is given by $\displaystyle y.$ The "height" of each shell (horizontal distance from side to side) I have explained in my previous post. It should be clear from looking at a picture.

Re: Need help with finding a volume of revolution with Cylindrical Shells method