Please Help. (1 Viewer)

[rr.dun.dun]

Does anyone know how to work out these questions?

Rosemary is having a party and decides to make an 'ice bowl' to hold punch. She freezes water in the space between two hemispherical bowls of diameter 16cm and 22cm.

(a) What is the volume of each ice bowl?
(b) What is the volume of the 'ice bowl'?
(c) What is the total surface area of the 'ice bowl'?
(d) How many litres of punch will the 'ice bowl' hold?

 

[rr.dun.dun]

anyone?

 

[yosemite sam]

does a) mean what is the volume of each bowl used? if it does...
a) V=(4/3 x pi x r^2) divided by 2
so for the 16cm diameter bowl, V=(4/3 x pi x 8^2) x0.5
= 134.04 cm^3 (2 d.p)
and for the 22cm diameter bowl, V=(4/3 x pi x 11^2) x 0.5
= 253.42cm^3 (2 d.p)
b) volume of the ice bowl would be the volume of the bigger bowl minus the volume of the smaller bowl
so, 253.42-134.04 = 119.38 cm^3 (2 d.p)
c)Total surface area would be half the surface area of the small bowl, half the surface area of the big bowl and then the annulus
so, formula for surface area is 4 x pi x r^2
Small bowl = (4 x pi x 8^2) divided by 2
=402.12cm2 (2. dp)
large bowl = (4 x pi x 11^2) divided by 2
= 760.27cm2 (2 d.p)
annulus = pi x 11^2 - pi x 8^2
= 121 - 64
= 57cm2
total surface area = 402.12+760.27+57
=1219.39cm2 (2 d.p)
d) volume in litres - cant think of it right now.
i'm not entirely sure if what ive done is right, comeone can feel free to correct it if its wrong. hope that helps

 

[LoneShadow]

looks correct to me. Assuming the smaller bowl is thin, the volume "ice bowl" can hold is the same as the volume of smaller bowl.

[of course in all of these we are assuming the top of the bowls used are in the same plane, else it has to be factored in]

 

[rr.dun.dun]

does a) mean what is the volume of each bowl used? if it does...
a) V=(4/3 x pi x r^2) divided by 2
so for the 16cm diameter bowl, V=(4/3 x pi x 8^2) x0.5
= 134.04 cm^3 (2 d.p)
and for the 22cm diameter bowl, V=(4/3 x pi x 11^2) x 0.5
= 253.42cm^3 (2 d.p)
b) volume of the ice bowl would be the volume of the bigger bowl minus the volume of the smaller bowl
so, 253.42-134.04 = 119.38 cm^3 (2 d.p)
c)Total surface area would be half the surface area of the small bowl, half the surface area of the big bowl and then the annulus
so, formula for surface area is 4 x pi x r^2
Small bowl = (4 x pi x 8^2) divided by 2
=402.12cm2 (2. dp)
large bowl = (4 x pi x 11^2) divided by 2
= 760.27cm2 (2 d.p)
annulus = pi x 11^2 - pi x 8^2
= 121 - 64
= 57cm2
total surface area = 402.12+760.27+57
=1219.39cm2 (2 d.p)
d) volume in litres - cant think of it right now.
i'm not entirely sure if what ive done is right, comeone can feel free to correct it if its wrong. hope that helps

the answer of (c) is wrong. the answer is supposed to be 1341cm2
why did you need the annulus?

 

[yosemite sam]

because if its the surface area then you would have the area on the outside of the bowl, the area on the inside and there would be a rim between them of a 3 cm annulus. like the top of a donut. if a donut was a bowl.

 

[Freshmex]

I'll finish it:

annulus = pi x 11^2 - pi x 8^2
= 121 - 64
= 57cm2
total surface area = 402.12+760.27+57
=1219.39cm2 (2 d.p)

The annulus * 57; it equals 179.07.
Otherwise, 402.12+760.27+179.07= 1341cm2.

As for D,

V(sphere)= 4/3πr^3. The answer you want is half the volume of the bowl's inner sphere; i.e. the sphere with a radius of 8cm: it is in this space which liquid can be held. So, 4/3π8^3= 2144.66cm^3, giving about 2.14 Litres of punch (neglecting that punch will occupy more space than water). That should be the answer.

 

[rr.dun.dun]

Thank You