A bowl is shaped like a hemisphere with diameter 30 cm. A ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of h centimeters.

Question

(a) Find the volume (in cubic centimeters) of water in the bowl if the ball is not completely submerged (0≤h≤10) . Volume = cubic centimeters (b) Find the volume (in cubic centimeters) of water in the bowl if the ball is completely submerged (10≤h≤15) . Volume = cubic centimeters

Answer 1

for part b.

case1. when water at at 10 cm exactly

ball volume of (4/3)π*5³

bowl volume of water (without ball) at 10 cm

is (using shell method and dx)

the integral of 2πx(15²-x²)^(1/2)dx from x=0 to x=10√2

=(2/3)π15³-(2/3)π(10√2)^(3/2)

the limit of x=10√2mis from y=15-h (y=5) and x=√(15²-y²)

when h=10, water volume V=[(2/3)π15³-(2/3)π(10√2)^(3/2)]-(4/3)π5³

case2. as h goes from 10 to 15

water is added in the amount of

integral 2πx(15²-x²)^(1/2)dx from 10√2 to x=√(30h-h²)^(1/2)

the formula for x limit is from y=15-h

(15-h)=(15²-x²)^(1/2),,,square both sides and solve for x

for part a.

case1. 0<h≤5

volume water in bowl without ball is integral 2πx(15²-x²)^(1/2)dx x from 0 to (30h-h²)^(1/2)

ball volume= integral 2πx(5²-x²)^(1/2)dx x from 0 to (10h-h²)^(1/2)

subtract ball from water volume for water with ball immersed volume

limit for x in ball volume from y=5-h,,,(5-h)²=5²-x²

case2. 5<h<10

volume water without ball is integral 2πx(15²-x²)^(1/2)dx x from 0 to (30h-h²)^(1/2)

ball volume= volume bottom half of ball (=(2/3)π5³)

plus volume formed on top half of ball = integral 2πx(5²-x²)^(1/2)dx x from (10h-h²)^(1/2) to 5

notice 5>(10h-h²)^(1/2) for h>5 so dx is increasing and integral value will be positive

again must subtract ball from water volume for water with ball immersed volime

Answer 2

V_bowl = (2/3)(15 - h)²h

V_ball = (4/3)(5 - h)²h

Subtract bottom equation from top.