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A Hemisphere Problem

Hi

I tried to calculate the surface area for this hemisphere but somehow it's wrong:

I used the formula for the surface area of a sphere which is 4pir^2 then divided it by 2 since it's an open top hemisphere. Here's my working

It has radius of 6 cm.

SO,

4pir^2
4*pi*6^2
=452.39
=452.39/2
=226.2cm

Why is this wrong?

Reply 1

DeanK2

A hemisphere also has an area on the bottom, not just the sides.

Reply 2

div curl F = 0

9

Well a hemisphere has a flat side as well as its curved side. The flat side is just a circle, so pi r^2.

So total surface area of a hemisphere is

3 \pi r^2

The OP said it was an open top hemisphere so it only has a curved surface area and her working is correct.

(Well I don't much like the use of the = sign as she is saying 452.39 = 226.2 but other than that it is OK)!

Reply 4

Rocious

12

DeanK2

A hemisphere also has an area on the bottom, not just the sides.

I interpreted open-top as meaning it didn't have the base thing.

So you'd multiply 2pir^2 by 2 cause of the inside?

Reply 5

DeanK2

Rocious

I interpreted open-top as meaning it didn't have the base thing.

So you'd multiply 2pir^2 by 2 cause of the inside?

If you did that you would have A of a sphere of radius 6. However, in all questions I assume surface area to be the area of a closed surface - i.e. the top or bottom.

DeanK2

If you did that you would have A of a sphere of radius 6. However, in all questions I assume surface area to be the area of a closed surface - i.e. the top or bottom.

Not necessarily! Using Rocious' reasoning a hollow sphere would have an area of

8\pi r^2

! I think we have to stick to the exterior surface area. Could the OP tell us why she thinks her answer is incorrect?

Reply 7

The Bachelor

13

Mr M

Not necessarily! Using Rocious' reasoning a hollow sphere would have an area of

8\pi r^2

!

I think this is the right answer. (

4 \pi r^2

) But, again, it really does depend on what one means by "open top".

The thing is, when we ask for the surface area of a shape, we are really talking about the external surface area. But once you take a hemisphere (and don't add a base to it), the area on the inside becomes part of the external surface area.

The Bachelor

I think this is the right answer. (

4 \pi r^2

) But, again, it really does depend on what one means by "open top".

The thing is, when we ask for the surface area of a shape, we are really talking about the external surface area. But once you take a hemisphere (and don't add a base to it), the area on the inside becomes part of the external surface area.

I have heard this argument before and I find it unconvincing. If the question had referred to the surface area of a hemispherical shell lamina, I might have agreed. How are you accounting for the ring around the top?

Reply 9

The Bachelor

13

Mr M

I have heard this argument before and I find it unconvincing. If the question had referred to the surface area of a hemispherical shell lamina, I might have agreed. How are you accounting for the ring around the top?

If you're talking about the ring "at the bottom of the hemisphere" which is a complete circumference, then that should have 0 area.

The Bachelor

If you're talking about the ring "at the bottom of the hemisphere" which is a complete circumference, then that should have 0 area.

And would you double the curved surface area of a cone as well or is that an exception to the rule?

Reply 11

The Bachelor

13

Mr M

And would you double the curved surface area of a cone as well or is that an exception to the rule?

If you're talking about a (closed) cone and the "spike" the place where the interior and exterior might intersect, I don't know. Is there a way to differentiate the interior and exterior, or not?

As a cone can be produced from a sector of a circle, it might make sense to consider just one surface. After all, you wouldn't propose that the sector is a lamina and then sum the areas of both sides. (Yes, I am aware of the 2D/3D distinction but I am playing Devil's Advocate).

The OP is a Year 11 GCSE student. I can confirm that if a GCSE student assumes a cone is infinitely thin and goes on to calculate the sum of the interior and exterior surface areas, he or she does not receive full marks.

Reply 13

The Bachelor

13

Mr M

As a cone can be produced from a sector of a circle, it might make sense to consider just one surface. After all, you wouldn't propose that the sector is a lamina and then sum the areas of both sides. (Yes, I am aware of the 2D/3D distinction but I am playing Devil's Advocate).

By similar argument, you can produce a torus by pasting a plane (of finite length and width) together; why do you not count the interior surface area, then?

Reply 14

Chewwy

17

The Bachelor

By similar argument, you can produce a torus by pasting a plane (of finite length and width) together; why do you not count the interior surface area, then?

your questions does not follow. Mr M wasn't saying you did. nobody ever said you counted the interior surface area of anything. because it's an utterly stupid and superfluous thing to do.

Reply 15

The Bachelor

13

Chewwy

your questions does not follow. Mr M wasn't saying you did. nobody ever said you counted the interior surface area of anything. because it's an utterly stupid and superfluous thing to do.