#### To determine

**To find:**

Volume of water in the bowl

#### Answer

i) Volume when the ball is partially immersed

V=10πh2

ii) Volume when ball is fully immersed

VT=13π45h2-h3-500 cm3

#### Explanation

**1) Formula:**

Volume of cap of sphere of radius r and height h is given by,

Vcaph=πh2r-h3= 13πh2(3r-h)

**2) Given:**

Hemisphere of diameter 30 cm

Heavy ball of diameter 10 cm

**3) Calculation:**

In order to calculate the volume of the water in the bowl, consider two cases- one is that the heavy ball is partially submerged and another is that the heavy ball is fully submerged.

Case 1:

Consider that water is filled upto height y in the bowl.

The marked region shown above is the bowl filled with water on the right side of the heavy ball, if rotated about y axis, will get the left side of bowl filled with water at h since the bowl is symmetric about y axis. So, the best method is to use the washer method.

So, area is calculated as

A=πouter radius2-πinner radius2

To calculate outer radius

x2=152-(15-y)2

x=152-(15-y)2

To calculate inner radius that is r

x2=52-(5-y)2

x=52-(5-y)2

Therefore,

A(y)=π152-(15-y)22-π52-(5-y)22

A(y)=π[152-15-y2]-π[52-5-y2]

A(y)=π[225-(225-30y+y2)]-π[25-(25-10y+y2)]

A(y)=π[225-225+30y-y2]-π[25-25+10y-y2]

A(y)=π[30h-y2-10y-y2]

A(y)=π[30y-y2-10y+y2]

A(y)=20πy

V=∫0h20πydy

V=20π∫0hydy

V=20π∫0hydy

V=20πy220h

V=10πh2

Case 2:

Suppose that the ball is completely immersed in water.

To find the volume of the water in the bowl simply subtract the volume of the ball from the volume of the bowl.

Volume of ball is equal to

V=43πr3=43πr53=500π3

According to formula of volume of cap,

At height h, the water is filled which can be calculated by the equation

Vcaph=πh2r-h3= 13πh2(3r-h)

Here r=15

Vcaph=13πh2(45-h)

Total volume is equal,

Vcap-Vball=13πh245-h-500π3

VT=13π45h2-h3-500 cm3

**Conclusion:**

Volume when the ball is partially immersed is

V=10πh2

Volume when the ball is fully immersed is

VT=13π45h2-h3-500 cm3