Finding a volume using the method of cylindrical shells which is generated by a parabola

Question

This problem is from the 7th edition of the book "Calculus and Analytic Geometry" by George Thomas and Ross Finney. It is problem number 3 of section 5.4

Problem:

Find the volume generated when the region bounded by the given curves and lines is revolved about the x-axis. (Note: $x = 0$ is the y-axis and $y = 0$ is the x-axis.) \begin{align*} y &= 3x - x^2 \\ y &= x \end{align*}

Answer:

I am going to use the method of cylindrical shells to compute the volume. Let $V$ be the volume we are asked to find.

The general form for $V$ when we are revolving about the x-axis is: $$ V = \int_a^b 2 \pi x f(x) \,\, dx $$ where $2 \pi x$ represents the circumference of the region. Since we are going around the x-axis $a = 0$. To find $b$ be we set up the following equation: \begin{align*} 3x - x^2 &= x \\ 2x - x^2 &= 0 \\ x^2 - 2x &= 0 \\ x(x-2) &= 0 \\ x = 0 \,&\text{ or } \, x = 2 \\ V &= \int_0^2 2 \pi x ( 3x - x^2 - x) \,\, dx \\ V &= 2 \pi \int_0^2 x ( 2x - x^2 ) \,\, dx \\ \int_0^2 x ( 2x - x^2 ) \,\, dx &= \int_0^2 2x^2 - x^3 \,\, dx \\ \int_0^2 x ( 2x - x^2 ) \,\, dx &= \frac{2x^3}{3} - \frac{x^4}{4} \Big|_0^2 \\ \int_0^2 x ( 2x - x^2 ) \,\, dx &= \frac{2(8)}{3} - \frac{16}{4} = \frac{16}{3} - 4 \\ \int_0^2 x ( 2x - x^2 ) \,\, dx &= \frac{4}{3} \\ % V &= 2 \pi \left( \frac{4}{3} \right) \\ V &= \frac{8 \pi}{3} \end{align*}

The book's answer is: $\frac{56\pi}{15}$

Where did I go wrong?

Answer 1

You said that you would use shells but you have actually used washers. The volume should be $$V=\int_0^2 2\pi y f(x) dx$$ Here $f(x)=(3x-x^2)-x$ is the length of the rotated vertical element and $y=\frac12 ((3x-x^2)+x)$ is the distance of the centroid (= mid-point) of the rotated element from the $x$ axis.

If you were to use shells then the volume would be $$V=\int_0^{9/4} 2\pi y g(y) dy$$ This is more difficult. The upper limit of integration is the peak of the parabola. $y$ is the distance of the rotated horizontal element from the $x$ axis. $g(y)$ is the length of the rotated horizontal element. It is a compound function. For $y=0..2$ it spans between the left side of the parabola and the straight line : $g(y)=x_0-x_1$ where $x_0=y$ and $x_1$ is the smaller solution of $x^2-3x+y=0$. For $y=2..9/4$ it spans between the left and right sides of the parabola : $g(y)=x_2-x_1$ where $x_2$ is the larger solution of $x^2-3x+y=0$.