Chapter 7.2, Problem 22E

To determine

To calculate:

To evaluate the given integral ${\displaystyle \int \frac{9r^{2}dr}{\sqrt{1-r^3}}}$ with the initial condition $u=1-r^3$

Answer to Problem 22E

The solution is $-6\sqrt{1-r^3}+c$

Explanation of Solution

Given information:

The given integral is ${\displaystyle \int \frac{9r^{2}dr}{\sqrt{1-r^3}}}$ with the initial condition $u=1-r^3$

Consider the initial condition $u=1-r^3$ ,

Differentiating both sides with respect to x,

  $\frac{du}{dr}=0-3r^2=-3r^2$

Since the integral has $r^{2}dr$ , solve this for $r^{2}dr$ ,

  $\begin{array}{l}\frac{du}{dr}=-3r^2\\ du=-3r^{2}dr\\ -\frac{1}{3}du=r^{2}dr\end{array}$

Substitute $u=1-r^3$ and $r^{2}dr=-\frac{1}{3}du$ into the integral ${\displaystyle \int \frac{9r^{2}dr}{\sqrt{1-r^3}}}$ ,

  ${\displaystyle \int \frac{9r^{2}dr}{\sqrt{1-r^3}}}={\displaystyle \int \frac{9\left(-\frac{1}{3}du\right)}{\sqrt{u}}}$

  $\begin{array}{l}={\displaystyle \int \frac{-3du}{\sqrt{u}}}\\ =-3{\displaystyle \int u^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}du}\end{array}$

On integrating we get,

  $\begin{array}{l}=-3\left(\frac{u^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)+c\\ =-6u^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+c\\ =-6\sqrt{u}+c\end{array}$

Now substituting back for u,

  $=-6\sqrt{1-r^3}+c$

Therefore, the solution is $-6\sqrt{1-r^3}+c$