Question

The volume of a sphere can be determined by the formula V = 4/3 πr^3, where r is the radius. What is the volume of a sphere with radius 3/2x^3y^–2 in terms of x and y?

Answer 1

Answer:

$V = \frac{99}{7} x^{9}y^{-6}$

Step-by-step explanation:

Given

$V = \frac{4}{3}\pi r^3$

Required

Find V when

$r = \frac{3}{2}x^3y^{-2$

Substitute $\frac{3}{2}x^3y^{-2$ for r in $V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi * (\frac{3}{2}x^3y^{-2})^3$

Open bracket

$V = \frac{4}{3}\pi * \frac{3}{2}^3 * x^{3*3}y^{-2*3}$

$V = \frac{4}{3}\pi * \frac{27}{8} * x^{9}y^{-6}$

Simplify the fractions

$V = \pi * \frac{9}{2} * x^{9}y^{-6}$

Let

$\pi = \frac{22}{7}$

So:

$V = \frac{22}{7} * \frac{9}{2} * x^{9}y^{-6}$

$V = \frac{11}{7} * \frac{9}{1} * x^{9}y^{-6}$

$V = \frac{99}{7} * x^{9}y^{-6}$

$V = \frac{99}{7} x^{9}y^{-6}$