# Volume of Solid of Revolution and Method of Cylinders

Sometimes it is very hard to use Method of Disks/Rings to obtain volume of solid of revolution.

But that's not a problem. We can use any shape for the cross section as long as it can be expanded or contracted to completely cover the solid.

In this note we introduce **method of cylindrical shells.**

First suppose that we rotate about y-axis. Imagine a hollow cylinder whose wall has width `Delta x`. In other words, imagine a rectangle of with `Delta x` rotated around y-axis. Radius of i-th cylinder is `r(x_i^**)` (it depends on `x_i^**`)where `x_i^**` lies in interval `[x_(i-1),x_i]` . On the figure radius is `r(x_i^**)=x_i^**`.

Imagine this shell to be cut and flattened. The resulting rectangular slab has length `2pir(x_i^**)` (length of circle with radius `r(x_i^**)`), width `Delta x` and height `h` which depends on `x_i^**` (on figure height is `f(x_i^**)`).

So, the volume of the shell is `2pir(x_i^**)h(x_i^**)Delta x`.

If we do this for every subinterval and add the results, we get an approximation to the volume of the solid: `V~~sum_(i=1)^n 2pir(x_i^**)h(x_i^**)Delta x`.

This approximation improves as increases, so `V=lim_(n->oo)sum_(i=1)^n 2pir(x_i^**)h(x_i^**)Delta x`.

We recognize in this limit of Riemann sum definite integral, therefore `V=int_a^b 2pir(x)h(x)dx`.

Volume of solid rotated around y-axis is `V=int_a^b 2pi r(x)h(x)dx`.

As can be seen the formula for area of the cross section in case of cylindrical shells is `A=2pi(radius)(height)`.

It is always a good practice to draw a sketch of problem in order to correctly determine radius and height.

In case we rotate around x-axis, radius and height will depend on `y`.

**Example 1.** Find the volume of the solid obtained by rotating about the y-axis the region bounded by the curve `y=4(x-1)^2(x-3)^2` and x-axis.

First of all note that we can't use method of disks here. Since we rotate about y-axis we need function in terms of `y`, but it is very hard to express `x` in terms of `y`.

Now we draw a sketch. In this case radius of cylinder is `x` and height is `y(x)=4(x-1)^2(x-3)^2`.

Bounds of integration are 1 and 3.

Therefore, `V=int_1^3 2pi x*4(x-1)^2(x-3)^2dx=`

`=8pi int_1^3 (x^4-8x^3+22x^2-24x+9)dx=`

`=8pi (1/5 x^5-2x^4+22/3 x^3-12x^2+9x)|_1^3=`

`=8pi (1/5*3^5-2*3^4+22/3*3^3-12*3^2+9*3)-8pi (1/5*1^5-2*1^4+22/3*1^3-12*1^2+9*1)=(128pi)/15`.

Next examples can be solved using method of disks, but we will solve them using cylindrical shells.

**Example 2.** Find the volume of the solid obtained by rotating about the y-axis the region bounded by the curves `y=sqrt(x)` and `y=x/2`.

First we draw a sketch. In this case radius of cylinder is `x` and height is difference between function values at `x`: `h(x)=sqrt(x)-x/2`.

Bounds of integration are points of intersection of `y=sqrt(x)` and `y=x/2`, i.e. 0 and 4.

Therefore, `V=int_0^4 2pi x*(sqrt(x)-x/2)dx=2pi int_0^4 (x^(3/2)-1/2 x^2)dx=`

`=2pi (2/5 x^(5/2)-1/6 x^3)|_0^4=2pi (2/5 *32 -64/6)=(64pi)/15`.

**Example 3.** Find the volume of the solid obtained by rotating about the line `x=-1` the region bounded by the curves `y=sqrt(x)` and `y=x/2`.

First we draw a sketch. In this case radius of cylinder is not `x`, it is distance between line `x=-1` and edge of cylinder: `1+x`. Height is `y(x)=sqrt(x)-x/2`.

Bounds of integration are points of intersection of `y=sqrt(x)` and `y=x/2`, i.e. 0 and 4.

Therefore, `V=int_0^4 2pi (1+x)*(sqrt(x)-x/2)dx=`

`=2pi int_0^4 (x^(3/2)+x^(1/2)-1/2 x^2 -1/2 x)dx=`

`=2pi (2/5 x^(5/2) +2/3 x^(3/2) -1/6 x^3-1/4 x^2)|_0^4=2pi (64/5+16/3-64/6-16/4)=(104pi)/15`.

Now let's see how to handle situations when we rotate about x-axis.

**Example 4.** Find the volume of the solid obtained by rotating about the x-axis the region bounded by the curve `y=x^2` on interval `[0,1]`.

Since we rotate about x-axis, we need function in terms of `y`: `x=+-sqrt(y)`. Since we are in first quadrant (interval `[0,1]`) then we choose positive value `x=sqrt(y)`.

Now we draw a sketch. In this case radius of cylinder is `y` and height is distance from 1 to function, i.e. `h(y)=1-sqrt(y)`.

Bounds of integration are 0 and 1.

Therefore, `V=int_0^1 2pi y*(1-sqrt(y))dy=`

`=2pi int_0^1 (y-y^(3/2))dy=2pi (1/2 y^2 -2/5 y^(5/2))|_0^1=2pi(1/2-2/5)=pi/5`.

**Example 5.** Find the volume of the solid obtained by rotating about the x-axis the region bounded by the curves `y=x^2` and `y=x`.

Since we rotate about x-axis, we need functions in terms of `y`: `x=sqrt(y)` and `x=y`.

Now we draw a sketch. In this case radius of cylinder is `y` and height is difference between outer and inner function at `y`: `h(y)=sqrt(y)-y`.

Bounds of integration are 0 and 1.

Therefore, `V=int_0^1 2pi y(sqrt(y)-y)dy=`

`=2pi int_0^1 (y^(3/2)-y^2)dy=2pi (2/5 y^(5/2)-1/3 y^3)|_0^1=2pi (2/5-1/3)=(2pi)/15`.

**Example 6.** Find the volume of the solid obtained by rotating about horizontal line `y=2` the region bounded by the curves `y=x^2` and `y=x`.

We again need functions in terms of `y`: `x=sqrt(y)` and `x=y`.

Bounds of integration are points of intersection of `x=sqrt(y)` and `x=y`, i.e. 0 and 1.

Now we draw a sketch.

At `y` radius of cylinder is `2-y` and height is difference between values of outer and inner functions at `y`: `sqrt(y)-y`.

Therefore, `V=int_0^1 2pi (2-y)(sqrt(y)-y)dy=2pi int_0^1 (2y^(1/2)+y^2-2y-y^(3/2))dy=`

`=2pi (4/3 y^(3/2)+1/3 y^3-y^2-2/5 y^(5/2))|_0^1=2pi(4/3 +1/3 -1-2/5)=(8pi)/15`.