Thread: SA and volume of half a sphere?

Originally Posted by xanith_47

Hello,
I need help with volume and surface areas of half a sphere (hemispheres)
I know that v= 4/3x pi x r cubed
SA= 4x pi x r squared

However, when I do this for half a sphere, i time the answer by half and when i refer to the answers at the back of my textbook, its different. Im not sure if the sphere is open or closed, since the book doesnt state that, and are there different formulas for finding out SA and V of half open and closed spheres?

Any help is greatly appreciated.
Thanks in advance

You mentioned those for a sphere, so half of those should be for a hemisphere.
If your answers and those at the back of the book are not the same, it could be that you misunderstood the problems so you could have used wrong formulas.
Can you post a question, with your answer and that answer at the back of the book? Let us investigate.

Calculate the surface area of the following sphere to the nearest m squared:

A hemisphere with diameter of 17m
The textbook doesnt mention whether its open or closed..if that makes a difference..

The answer at the back is 681m squared.... and i got 453.96m squard= 454m squard. I used 2 x pi x 8.5 squard = formula for the SA of half a sphere.

Can you explain to me how this works?

Originally Posted by xanith_47

Calculate the surface area of the following sphere to the nearest m squared:
A hemisphere with diameter of 17m
The textbook doesnt mention whether its open or closed..if that makes a difference..
The answer at the back is 681m squared.... and i got 453.96m squard= 454m squard. I used 2 x pi x 8.5 squard = formula for the SA of half a sphere.
Can you explain to me how this works?

Hello,

your result is OK - and the result in the book is OK too.

The book defines the surface of a hemisphere as the half of a complete sphere plus the area of the covering circle .

So you get:

Greetings

EB

Umm.. so would the "the area of the covering circle" be the area INSIDE the circle? How come then, when the hemisphere is closed, we dont count that area which closes the circle, but we count the area inside it when its open. So your saying that the extra area inside the circle= pi x r squard= area of a normal circle?

Thanks in advance

Originally Posted by xanith_47

Umm.. so would the "the area of the covering circle" be the area INSIDE the circle? How come then, when the hemisphere is closed, we dont count that area which closes the circle, but we count the area inside it when its open. So your saying that the extra area inside the circle= pi x r squard= area of a normal circle?

Thanks in advance

Hello,

I've attached a diagram to show you where you find this additional area.

Greetings

EB

Ok thanks for the help.. so if there was no covering circle (hence its open), then we dont calculate the area inside the circle..just the exterior area? Thanks

Originally Posted by xanith_47

Ok thanks for the help.. so if there was no covering circle (hence its open), then we dont calculate the area inside the circle..just the exterior area? Thanks

My half-cent.....
Supposed to be, a sphere is solid. It is not hollow--unless it is specified to be hollow.
Like a cube. You won't think of a cube as hollow if all is said is it is a cube.

So the hemisphere in the problem is supposed to be solid. Nothing was mentioned that it was hollow or solid.
Hence its surface area is the sum of the areas of the curved surface of the dome and of the flat surface of the base.
The SA of the dome is (1/2)[4pi(r^2)]
The base is a great circle of the sphere where the hemisphere came from (a great circle of a sphere has the center of the sphere as its center also). Its SA is pi(r^2).