Q. 17 - Calculus

Question

Write the volumes of the solids of revolution shown in Exercises 17–20 in terms of definite integrals that represent accumulations of shells. Do not solve the integrals.

Step-by-Step Solution

Answer

The volume of the solid in terms of definite integral is $2\mathrm{\pi }{\int }_0^3\mathrm{y}\left(3-{\mathrm{f}}^{-1}\left(\mathrm{y}\right)\right)d\mathrm{y}$.

Step 1. Given information.

Consider the given figure that represent accumulations of shells.

Step 2. Find the volume of solid in terms of definite integral.

The function is terms of y is:

$$\begin{gathered} y=f\left(x\right) \\ x=f^{-1}\left(y\right) \end{gathered}$$

From $y=0$ to $y=2$, the height of the shell at $y_k^{*}$ will be the difference $\left(3-y_k^{*}\right)$.

So, the volume of the solid obtained by revolving the function about the x-axis shown in the figure is:

$V=2\mathrm{\pi }{\int }_0^{2}y\left(3-f^{-1}\left(y\right)\right)dy$