# Volume by Cylindrical Shells Method

Find the volume of a solid of revolution generated by revolving a region bounded by the graph of a function around one of the axes using definite integrals and the method of cylindrical shells where the integration is perpendicular to the axis of rotation. Exercises with their answers is presented at the bottom of the page.

## Formula - Method of Cylindrical Shells

If \( f \) is a function such that \( f(x) \ge 0 \) (see graph on the left below) for all \( x \) in the interval [ \( x_1 \), \( x_2 \) ], the volume of the solid generated by revolving, around the \( y \) axis, the region bounded by the graph of \( f \), the \( x \) axis (\( y = 0 \)) and the vertical lines \( x = x_1 \) and \( x = x_2 \) is given by the integralIf \( g \) is a function such that \( g(y) \ge 0 \) for all \( y \) in the interval [ \( y_1 \), \( y_2 \) ], the volume of the solid generated by revolving, around the \( x \) axis, the region bounded by the graph of \( g \), the \( y \) axis (\( x = 0 \)) and the horizontal lines \( y = y_1 \) and \( y = y_2 \) is given by the integral

\[ \Large {\text{Volume} = \color{red}{\int_{y_1}^{y_2} 2\pi \; y \; g(y) \; dy}} \]

## Examples with Detailed Solutions

### Example 1

Use the method of cylindrical shells to find the volume of the solid generated by revolving the shaded (red) region (triangle) about the \( y \) axis.### Solution to Example 1

Note that this problem has been solved in Volume of a Solid of Revolution using the washers method. let's now solve it using the cylindrical shells method and you may compare the two methods.The shape to be rotated is limited by \( x = 0 \), \( y = 0 \) and two curves \( y = x \) and \( y = - x + 2 \) with a point of intersection at (1,1) and we therefore need to split the integral to find the volume into two parts as follows:

\[ \text{Volume} = \int_{x_1}^{x_2} 2\pi x f(x) dx \]

Split the interval of integration

\[ = \int_{0}^{1} 2\pi x (x) dx + \int_{1}^{2} 2\pi x (-x + 2) dx \]

Integrate

\[ = 2\pi \left[x^3/3\right]_0^1 + 2\pi\left[-x^3/3 + x^2\right]_1^2 \]

Evaluate

\[ = 2\pi/3 + 2\pi \left[-8/3+4 -(-1/3+1) \right] \]

Simplify

\[ = 2\pi/3 + 2\pi \left[4/3 - 2/3 \right] \]

\[ = 2\pi/3 + 4\pi/3 = 2\pi \]

### Example 2

Use the method of cylindrical shells to find the volume of the solid generated by revolving the area enclosed by \( y = - x^3 + 2 x^2 - x + 2 \) and \( y = -x + 1 \) in the first quadrant.### Solution to Example 2

The graphs of \( y = - x^3 + 2 x^2 - x + 2 \) and \( y = -x + 1 \) are shown below. Both graphs have \( x \) intercepts calculated by solving the equations \( y = 0 \).\( x \) intercept of \( y = - x^3 + 2 x^2 - x + 2 \) is found by solving

\( - x^3 + 2 x^2 - x + 2 = 0 \)

Factor the left side of the above equation

\( x^2(- x + 2) +(- x + 2) = 0 \)

\( (- x + 2) (1 + x^2) = 0 \)

\( - x + 2 = 0 \) gives the \( x \) intercept at \( x = 2 \) as shown in the graph above.

\( 1 + x^2 = 0 \) has no real solutions

\( x \) intercept of \( y = - x + 1 \) is found by solving the equation:

\( - x + 1 = 0 \)

whose solution is \( x = 1 \) a shown in the graph above. Note that from \( x = 0 \) to \( x = 1 \), the upper part of the region is limited by \( y = f(x) = - x^3 + 2 x^2 - x + 2 \) and the lower part is limited by \( y = h(x) = - x + 1 \)

The volume \( V_1 \) from \( x = 0 \) to \( x = 1 \) is given by

\[ V_1 = \int_{0}^{1} 2\pi x (f(x) - h(x)) dx \]

\[ = 2\pi \int_{0}^{1} x (- x^3 + 2 x^2 - x + 2 - (-x+1)) dx \]

\[ = 2\pi \int_{0}^{1} (- x^4 + 2 x^3 + x) dx \]

\[ = 2\pi \left[ - x^5/5 + 2 x^4/4 + x^2/2 \right]_0^1 = 8\pi / 5\]

From \( x = 1 \) to \( x = 2 \), the upper part of the region is limited by \( y = f(x) = - x^3 + 2 x^2 - x + 2 \) and the lower part is limited by \( y = 0 \), hence the volume \( V_2 \) is given by \[ V_2 = \int_{1}^{2} 2\pi x f(x) dx \]

\[ = 2\pi \int_{1}^{2} x (- x^3 + 2 x^2 - x + 2) dx \]

\[ = 2\pi \int_{1}^{2} (- x^4 + 2 x^3 - x^2 + 2 x) dx \]

\[ 2\pi \left[ - x^5/5 + 2 x^4/4 - x^3/3 + 2 x^2/2 \right]_1^2 = 59\pi/15\] The total volume \( V \) is the sum of the two volumes found above

\[ V = V_1 + V_2 = 8\pi / 5 + 59\pi/15 = 83 \pi / 15 \]

### Example 3

Find a formula for the volume of the solid generated by revolving the area enclosed by \( y = 0 \), \( x = 0 \) and \( y = a + \sin(x) \)for \( 0 \leq x \leq 2\pi \) and \( a \geq 1 \) about the y-axis.

### Solution to Example 3

The volume of the solid of revolution described in this example is given by\[ V = \int_{0}^{2\pi} 2\pi x ( a + \sin(x) ) dx = 2\pi \int_{0}^{2\pi} a x dx + 2\pi \int_{0}^{2\pi} x \sin (x) dx \] Calculate the two integrals separately \[ I_1 = 2\pi a \int_{0}^{2\pi} x dx = 2\pi a \left [ x^2/2 \right ]_0^{2\pi} = 4\pi^3 a \] \[ I_2 = 2\pi \int_{0}^{2\pi} x \sin(x) dx \] Let \( U = x \) and \( V' = \sin(x) \) and use integration by parts to obtain \[ I_2 = 2\pi[ - x \cos(x)]_{0}^{2\pi} + 2\pi \int_{0}^{2\pi} \cos(x) dx \]

\[ = 2\pi\left[ - x \cos(x)\right]_{0}^{2\pi} +2\pi[ \sin(x) ]_{0}^{2\pi} = -4\pi^2 \] Total volume \( V \) is equal to \[ V = I_1 + I_2 = 4\pi^3 a -4\pi^2 \]

### Example 4

Use the method of method of cylindrical shells to find a formula for the volume of the solid generated by revolving the area enclosed by \( y = 0 \), \( x = 0 \) and \( \left(\dfrac{x}{a}\right)^2 + \left(\dfrac{y}{b}\right)^2 = 1 \) in the first quadrant about the x-axis (\( a \) and \( b \) both positive, )### Solution to Example 4

We first solve for \( x \) to find the equation of the curve \( \left(\dfrac{x}{a}\right)^2 + \left(\dfrac{y}{b}\right)^2 = 1 \) in the first quadrant (\( x > 0 \) and \( y > 0 \))\( x = a \sqrt{1 - \left(\dfrac{y}{b}\right)^2} \)

The rotation is around the \( x \) axis therefore the cylindrical shells are parallel to the \( x \) axis and the volume \( V \) is given by

\[ V = \int_{0}^{b} 2\pi y \left( a \sqrt{ 1 - \left(\dfrac{y}{b}\right)^2} \right) dy \] Let us use the substitution \( u = 1 - \left(\dfrac{y}{b}\right)^2 \) which gives \( \dfrac{du}{dy} = -\dfrac{2 y}{b^2} \), \( u = 1 \) when \( y = 0 \) and \( u = 0 \) when \( y = b \), hence after substitution \( V \) is given by \[ V = 2\pi a \left( -\dfrac{b^2}{2} \right) \int_{1}^{0} u^{\dfrac{1}{2}} du \]

\[ = - \pi a b^2 \left( \dfrac{2}{3} \right) u^{\dfrac{3}{2}} \Bigg|_{1}^{0} \]

\[ = \dfrac{2\pi a b^2}{3} \]

## Exercises

(1) Find the volume of the solid generated when part the graph of \( f(x) = -x^4 + 3x^3 - x + 3 \) in quadrant (I) is revolved about the \( y \) axis . (Hint: graph \( f \) and find the \( x \) and \( y \) intercepts)### Answers to Above Exercises

(1) \( \dfrac{288\pi}{5} \)## More Links and References

Volume of a Solid of RevolutionArea under a curve .

Area between two curves .

integrals and their applications in calculus.