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

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Answer 1 · 2016-10-17T12:48:28

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

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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,

$8.5/2=4.25$

$"(As the radius is half the diameter)"$

Let's calculate

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)("Volume"$

$rarrcolor(blue)(4pir^2$

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

$rarrcolor(blue)(88/7*18.06$

$rarrcolor(blue)(12.57*18.06$

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

$color(white)(45$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

$rarrcolor(indigo)(4/3pir^3$

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

$rarrcolor(indigo)(88/21*76.76$

$rarrcolor(indigo)(4.19*76.76$

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

$color(white)(45$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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