How to 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? We will present examples based on the methods of disks and washers where the integration is parallel to the axis of rotation. A set of exercises with answers is presented at the end.

\[ \large \text{Volume} = \color{red}{\int_{y_1}^{y_2} \pi [ z(y)^2 - w(y)^2 ] \, dy} \]

We now use definite integrals to find the volume defined above. If we let \( f(x) = x \) according to formula 1 above, the volume is given by the definite integral \[ Volume = \int_{0}^{2} \pi x^2 dx = \left [\pi \dfrac{x^3}{3} \right ]_0^2 = \dfrac{8\pi}{3} \] The first method works because \( y = x \) is a linear function and the volume generated is that of a right circular cone , however the second method work for shapes other than cones and will be used in the examples below.

The graph of \( y = \sqrt{r^2 - x^2} \) is shown above and \( y \geq 0 \) from \( x = -r \) to \( x = r \). The volume is given by formula 1 as follows

\[ Volume = \int_{-r}^{r} \pi (\sqrt{r^2 - x^2})^2 dx = \int_{-r}^{r} \pi (r^2 - x^2) dx \] \[ = \pi\left [r^2 x - \dfrac{x^3}{3} \right ]_{-r}^r \] \[ = \pi\left [ (r^3 - \dfrac{r^3}{3}) - (-r^3 + \dfrac{r^3}{3}) \right ] \] \[ = \dfrac{4}{3} \pi r^3 \] This is the very well known formula for the volume of the sphere. If you revolve a semi circle of radius \( r \) around the x axis, it will generate a sphere of radius \( r \).

\[ Volume = \int_{0}^{1} \pi [ (-y+2)^2 - y^2] \, dy \] \[ = \int_{0}^{1} \pi [ - 4 y + 4] \, dy \\ = \pi \left [-2y^2+4y \right ]_0^1 = 2\pi \]

\[ Volume = \int_{0}^{2} \pi ( f(x)^2 - h(x)^2 ) \, dx \] \[ = \int_{0}^{2} \pi ( (1+x^2)^2 - (\sqrt x)^2 ) \, dx \] \[ = \pi \int_{0}^{2} ( 1+x^4+2x^2 - x ) \, dx \] \[ = \pi \left [ \dfrac{x^5}{5} + \dfrac{2x^3}{3} - \dfrac{x^2}{2} + x \right]_0^2 \] \[ = \dfrac{176\pi}{15} \]

\( y = R + \sqrt{r^2 - x^2} \) , upper semi circle , and \( y = R - \sqrt{r^2 - x^2} \) , lower semi circle

The torus is generated by rotating the two halves semi circles the x axis hence the use of formula 2 given above to find the volume of the torus. Let \( f(x) = R + \sqrt{r^2 - x^2} \) and \( h(x) = R - \sqrt{r^2 - x^2} \). Because of the symmetry of the circle and therefore the torus with respect to the y axis, we integrate from \( x = 0 \) to \( x = r \) then double the answer to find the total volume.

\[ Volume = \int_{0}^{r} \pi [ f(x)^2 - h(x)^2 ] \, dx \] \[ = \pi \int_{0}^{r} [ (R + \sqrt{r^2 - x^2})^2 - (R - \sqrt{r^2 - x^2})^2 ] \, dx \] \[ = 4 R \pi \int_{0}^{r} [ \sqrt{r^2 - x^2} ] \, dx \] \[ = 4 R \pi (\dfrac{1}{2}) \left [ x \sqrt{r^2-x^2} + r^2\arctan(\dfrac{x}{\sqrt{r^2-x^2}}) \right ]_0^r \] \[ = 4 R \pi ( \pi r^2 / 4) \] \[ = {\pi}^2 R r^2\] The total volume is twice the above, hence the volume of a torus is given by \[ Volume = 2 {\pi}^2 R r^2 \]

(2) Find the volume generated when the finite region bounded by the curves \( y = x^3 \), \( y = x^2 \) is revolved about the y axis.(hint: you need to find the points of intersections of the two curves)

(2) \( \dfrac{\pi}{10} \)

integrals and their applications in calculus.

Math Problems with Solutions