# What is the surface area and volume of a bowling ball with a diameter of 8.5 inches?

Oct 17, 2016

color(blue)("Volume"=227.01 color(blue)("in"^3$\mathmr{and}$ color(indigo)("Surface area"=321.62 color(indigo)("in"^2

#### Explanation:

We use these formulas to calculate the volume and surface area.
(Assume the bowling ball does not have a holes)

As, the bowling ball is a sphere.So, we use

color(blue)("Volume of the sphere"=4pir^2

color(indigo)("Surface area of sphere"=4/3pir^3

Where,

color(orange)(r="radius"

color(orange)(pi=22/7

We know that the diameter of the ball is $8.5$.So the radius will be,

$\frac{8.5}{2} = 4.25$

$\text{(As the radius is half the diameter)}$

Let's calculate

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color(blue)("Volume"

rarrcolor(blue)(4pir^2

rarrcolor(blue)(4*22/7*4.25^2

rarrcolor(blue)(88/7*18.06

rarrcolor(blue)(12.57*18.06

rArrcolor(green)(227.01 color(green)("in"^3

color(white)(45

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color(white)(5)color(indigo)("Surface area"

rarrcolor(indigo)(4/3pir^3

rarrcolor(indigo)(4/3*22/7*4.25^3

rarrcolor(indigo)(88/21*76.76

rarrcolor(indigo)(4.19*76.76

rArrcolor(green)(321.62 color(green)("in"^2

color(white)(45

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:.color(blue)("Volume"=227.01 color(blue)("in"^3$\mathmr{and}$ color(indigo)("Surface area"=321.62 color(indigo)("in"^2