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.

If \( g \) 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

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

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

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

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

\( x \) 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

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

for \( 0 \leq x \leq 2\pi \) and \( a \geq 1 \) about the y-axis.

\[ V = \int_{0}^{2\pi} 2\pi x ( a + \sin(x) ) dx = 2\pi \int_{0}^{2\pi} a x dx + 2\pi \int_{0}^{2\pi} x \sin (x) dx \] 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 \[ V = I_1 + I_2 = 4\pi^3 a -4\pi^2 \]

\( x = a \sqrt{1 - \left(\dfrac{y}{b}\right)^2} \)

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 \[ V = 2\pi a \left( -\dfrac{b^2}{2} \right) \int_{1}^{0} u^{\dfrac{1}{2}} du \]

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

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

Area under a curve.

Area between two curves.

integrals and their applications in calculus.