Find the Function V(r)=4/3pir^3

$V (r) = \frac{4}{3} π r^{3}$

The function $6 (r)$ can be found by evaluating the indefinite integral of the derivative $V (r)$ .

$6 (r) = \int V (r) d r$

Simplify.

Combine $π$ and $\frac{4}{3}$ .

$\int \frac{π \cdot 4}{3} \cdot r^{3} d r$

Combine $\frac{π \cdot 4}{3}$ and $r^{3}$ .

$\int \frac{π \cdot 4 r^{3}}{3} d r$

Move $4$ to the left of $π$ .

$\int \frac{4 π r^{3}}{3} d r$

Since $\frac{4 π}{3}$ is constant with respect to $r$ , move $\frac{4 π}{3}$ out of the integral.

$\frac{4 π}{3} \int r^{3} d r$

By the Power Rule, the integral of $r^{3}$ with respect to $r$ is $\frac{1}{4} r^{4}$ .

$\frac{4 π}{3} (\frac{1}{4} r^{4} + C)$

Simplify the answer.

Rewrite $\frac{4 π}{3} (\frac{1}{4} r^{4} + C)$ as $\frac{4 π}{3} \cdot \frac{1}{4} r^{4} + C$ .

$\frac{4 π}{3} \cdot \frac{1}{4} r^{4} + C$

Simplify.

Multiply $\frac{4 π}{3}$ by $\frac{1}{4}$ .

$\frac{4 π}{3 \cdot 4} r^{4} + C$

Multiply $3$ by $4$ .

$\frac{4 π}{12} r^{4} + C$

Cancel the common factor of $4$ and $12$ .

Factor $4$ out of $4 π$ .

$\frac{4 (π)}{12} r^{4} + C$

Cancel the common factors.

Factor $4$ out of $12$ .

$\frac{4 π}{4 \cdot 3} r^{4} + C$

Cancel the common factor.

$\frac{4 π}{4 \cdot 3} r^{4} + C$

Rewrite the expression.

$\frac{π}{3} r^{4} + C$

The function $6$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$6 (r) = \frac{π}{3} r^{4} + C$