Here's c by using washers (solid of revolution).
We revolve the upper half circle about the x-axis.
Since the upper half circle is
The volume of a sphere is:
hi how can the following be proved using integral methods:
a) prove surface area of sphere, radius a, is
b) prove area of a disk, radius a, is
c) prove volume of ball, radius a, is
d) prove volume of axisymmetric cone of height h and base with radius a, is
i think my working of (a) is correct:
working of (a):
use spherical co-ordinates:
|S| = = = =
how can i do the rest please. and what integration methods should I be using for each? thnx xxxx
No, I didn't skip anything. If one wanted to be anal, one could prove why solids of revolutions work. But that's a little much.
For the cone. You could put the point of the cone at the origin and find the equation of the line that makes up a side of the cone.
Solve for x and get
Now, revolve it about the y-axis and integrate from 0 to h:
For the area of a disk. Just use the equation of a circle centerd at the origin and integrate from -a to a.
You have the surface area of a sphere OK. But let's do it another show off way. Surface integrals.
If we project a hemisphere onto the xy plane, we get a circle with radius a. BUT, we can't aplly it directly because the partial derivatives don't exist on the boundary of the region since there.
So, we use a limit and evaluate over a wee tinier smaller region of radius, say, p and let r approach a.
Of course, that's for a hemisphere. Multiply by 2 for a sphere and get:
There. Wasn't that fun?.
thank you . i just read up about washers and it seems very easy to show the volumes etc using this method.
is my method undesirable? considering i used spherical co-oridnates and it was a PAINSTAKING process to work out the
this site seems to show the solid of revolution. but they seem to be more like formulas rather than proofs: Solid of Revolution -- from Wolfram MathWorld
so was your method simply a formula used.. or would it just as much be considered a proof?
lastly, is it possible to prove the volume is (4/3pi a^3) ... using spherical co-ordinates.. ? if so how would you start it off...
thanks appreciated very much
lol that proof you have for surface of sphere was quite extreme hehehe. wouldnt have been able to come up with it myself but i think i do understand which was helpful. cheers hun.
i'll try working out the rest the way you originally showed me. i'll show you my working later when i get back home, have to rush at the moment to see bf! thank u xx
i wanted to acutally know. is there a way to solve them using integral methods such as divergence, or greens, or stoke's??
or simply using spherical/polar/cartesian co-ordinates??
for example, i worked out (A) using spherical co-ordinates
where r = a
so then i used the parameter
now cross product of
the magnitude is therefore:
so that is how i managed to work out the surface area of the sphere.
do you have a similiar simple way of solving the others instead of washer's method and that extreme method you showed for the surface please? thnx xx
Here's the area of a circle of radius r using Green's theorem. If you look in calc texts on the subject, all they ever show the area of is an ellipse. I worked this up for a circle.
Green's theorem area formula:
This simplifies to:
This is area. I don't know about using it for volume. Maybe look into it.
thanks galactus. i actually managed to do the area of a circle myself aswell:
check this site out, it shows how to do it exactly the same way you did:
go right to the bottom of the page: example 4:
any suggestions on how to work out the volume of the ball and cone though?
presumably part (C) needs to use spherical polar coordinates and do a volume (triple) integral.
Part (d) seems to be Cylindrical polar coordinates with a volume integral.
but i don't know which theorem this would be and what the parameters would be? any suggestions please?
The volume of a sphere in spherical would be:
The volume of a cone in polar triple is:
This comes about by placing the cones 'point' on the origin. The equation of said cone in rectangular coordinates is
Since implies that
Therefore, the cone has equation in polar coordinates.
This gives the above integral which evaluates to
We have the circle: . . (upper semicircle)a) Prove the surface area of a sphere of radius is:
Area of a surface of revolution: .
We have: .
. . Then: .
So we have: .
We have the circle: . . (upper semicircle)b) Prove the area of a disk of radius is:
The area under the semicircle is: .
The area of the entire circle is: .
We have: .