# 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 ShellsIf f is a function such that f(x) ≥ 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 integral
If g is a function such that g(y) ≥ 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 1Use 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 1Note 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} = \int_{0}^{1} 2\pi x (x) dx} + \int_{1}^{2} 2\pi x (-x + 2) dx} \\\\
= 2\pi [x^3/3]_0^1 + 2\pi[-x^3/3 + x^2]_1^2\\\\
= 2\pi/3 + 2\pi \left [-8/3+4 -(-1/3+1) \right ] = 2\pi/3 + 2\pi \left [4/3 - 2/3 \right ] \\\\
= 2\pi/3 + 4\pi/3 = 2\pi
## Example 2Use 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 2The 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
- x ^{3} + 2 x^{2} - x + 2 = 0
Factor. 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 - x + 1 = 0 gives 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, hence the volume V_{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 3Find 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 ≤ x ≤ 2π and a ≥ 1 about the y-axis. ## Solution to Example 3The 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[ - x \cos(x)]_{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 4Use 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 (x/a)^{2} + (y/b)^{2} = 1 in the first quadrant about the x-axis (a and b both positive, )
## Solution to Example 4We first solve for x to find the equation of the curve (x/a)^{2} + (y/b)^{2} = 1 in the first quadrant (x > 0 and y > 0)x = a √(1 - (y/b) ^{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 ( a \sqrt{ 1 - (y/b)^2} ) dy
Let us use the substitution u = 1 - (y/b)^{2} which gives du/dy = -2 y /b^{2} , u = 1 when y = 0 and u = 0 when y = b, hence after substitution V is given hy
V = 2\pi a (-b^2/2) \int_{1}^{0} u^{1/2} du \\\\
= - \pi a b^2 \left [(2/3) u^{3/2} \right ]_1^0 \\\\
= \frac{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) 288π/5## More Links and ReferencesVolume of a Solid of RevolutionArea under a curve. Area between two curves. integrals and their applications in calculus. |