# Thread: areas, surfaces and volumes...

1. ## areas, surfaces and volumes...

hi how can the following be proved using integral methods:

a) prove surface area of sphere, radius a, is $4 \pi a^2$
b) prove area of a disk, radius a, is $\pi a^2$
c) prove volume of ball, radius a, is $\frac{4}{3} \pi a^3$
d) prove volume of axisymmetric cone of height h and base with radius a, is $\frac{1}{3}\pi a^2 h$

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i think my working of (a) is correct:
working of (a):
use spherical co-ordinates:
|S| = $\int\int_{D} ||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| dA$

$||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi$

so:
|S| = $\int^{2\pi}_{0}\int^{\pi}_{0} a^2sin \phi d\phi d\theta$ = $\int^{2\pi}_{0} \left[ -a^2cos\phi \right]^{\pi}_0 d\theta$ = $\int^{2\pi}_{0} 2a^2 d\theta$ = $4\pi a^2$

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how can i do the rest please. and what integration methods should I be using for each? thnx xxxx

2. Here's c by using washers (solid of revolution).

We revolve the upper half circle $x^{2}+y^{2}=a^{2}$ about the x-axis.

Since the upper half circle is $y=f(x)=\sqrt{a^{2}-x^{2}}$

The volume of a sphere is:

$\pi\int_{a}^{b}[f(x)]^{2}dx$

${\pi}\int_{-a}^{a}(a^{2}-x^{2})dx$

$\pi\left[a^{2}x-\frac{a^{3}}{3}\right]_{a=-r}^{r}=\frac{4}{3}{\pi}a^{3}$

3. thnx hun. would your proof of (c) that be considered a "complete proof" or did you skip out a lot of steps in order to achieve that solution?

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any suggestions for the other ones please?

xx

4. 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.

$y=\frac{h}{a}x$

Solve for x and get $x=\frac{a}{h}y$

Now, revolve it about the y-axis and integrate from 0 to h:

$\pi\int_{0}^{h}\left[\frac{a}{h}y\right]^{2}dy=\frac{1}{3}{\pi}a^{2}h$

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.

$z=\sqrt{a^{2}-x^{2}-y^{2}}$

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 $x^{2}+y^{2}=a^{2}$ there.

So, we use a limit and evaluate over a wee tinier smaller region of radius, say, p and let r approach a.

$\lim_{p\rightarrow{a^{-}}}\int\int\sqrt{(\frac{\partial{z}}{\partial{x}}) ^{2}+(\frac{\partial{z}}{\partial{y}})^{2}+1} \;\ dA$

= $\lim_{p\rightarrow{a^{-}}}\int\int\sqrt{\frac{x^{2}}{a^{2}-x^{2}-y^{2}}+\frac{y^{2}}{a^{2}-x^{2}-y^{2}}+1} \;\ dA$

= $\lim_{p\rightarrow{a^{-}}}\int\int\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}} \;\ dA$

= $\lim_{p\rightarrow{a^{-}}}\int_{0}^{2\pi}\int_{0}^{r}\frac{a}{\sqrt{a^{2}-r^{2}}}r \;\ dr \;\ d{\theta}$

= $\lim_{p\rightarrow{a^{-}}}\int_{0}^{2\pi}\left[-a\sqrt{a^{2}-r^{2}}\right]_{r=0}^{p} \;\ d{\theta}$

= $\lim_{p\rightarrow{a^{-}}}\int_{0}^{2\pi}\left(a^{2}-a\sqrt{a^{2}-p^{2}}\right) \;\ d{\theta}$

= $\lim_{p\rightarrow{a^{-}}}2{\pi}\left(a^{2}-a\sqrt{a^{2}-p^{2}}\right)=2{\pi}a^{2}$

Of course, that's for a hemisphere. Multiply by 2 for a sphere and get:

$4{\pi}a^{2}$

There. Wasn't that fun?.

5. 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

$||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi
$

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

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so was your method simply a formula used.. or would it just as much be considered a proof?
for example,

Originally Posted by galactus
We revolve the upper half circle $x^{2}+y^{2}=a^{2}$ about the x-axis.

Since the upper half circle is $y=f(x)=\sqrt{a^{2}-x^{2}}$

The volume of a sphere is:

$\pi\int_{a}^{b}[f(x)]^{2}dx$
the assumption made here was that $y=f(x)=\sqrt{a^{2}-x^{2}}$ can be used to show the volume of a sphere. wouldn't you actually need to prove assumptions made?

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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

6. No, why would you need to go to that extreme?.

We learn early on in math the equation of a circle.

Note, I updated my previous post for the surface area of a sphere using a surface integral.

7. 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

8. 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 $r(\theta,\phi) = (asin\phi cos\theta)i + (asin\phi sin\theta)j + (2cos\phi)k$

now cross product of $\frac{dr}{d\theta}X \frac {dr}{d\phi} = -(a^2sin^2\phi cos\theta)i - (a^2sin^2\phi sin\theta)j - (a^2sin\phi cos\phi)k$

the magnitude is therefore:

$||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi$

so:
$|S| = \int^{2\pi}_{0}\int^{\pi}_{0} a^2sin \phi d\phi d\theta = \int^{2\pi}_{0} \left[ -a^2cos\phi \right]^{\pi}_0 d\theta = \int^{2\pi}_{0} 2a^2 d\theta = 4\pi a^2$

so that is how i managed to work out the surface area of the sphere.
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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

9. 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: $\frac{1}{2}\oint_{C}[xdy-ydx]$

$c(t)=(rcos(t), rsin(t))$

$c'(t)=(-rsin(t),rcos(t))$

$\frac{1}{2}\int_{0}^{2\pi}\left[(rcos(t))(rcos(t))-(rsin(t))(-rsin(t))\right]dt$

This simplifies to:

$\frac{1}{2}\int_{0}^{2\pi}r^{2}dt={\pi}r^{2}$

Voila!.

This is area. I don't know about using it for volume. Maybe look into it.

10. 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:

http://tutorial.math.lamar.edu/Class...nsTheorem.aspx
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?

11. The volume of a sphere in spherical would be:

$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{a}({\rho}^{ 2}sin(\phi))d{\rho}d{\phi}d{\theta}$

The volume of a cone in polar triple is:

$\int_{0}^{2\pi}\int_{0}^{R}\int_{\frac{Hr}{R}}^{H} r \;\ dz \;\ dr \;\ d{\theta}$

This comes about by placing the cones 'point' on the origin. The equation of said cone in rectangular coordinates is

$z=\frac{H\sqrt{x^{2}+y^{2}}}{R}$

Since $z=H \;\ and \;\ \frac{H\sqrt{x^{2}+y^{2}}}{R}$ implies that $x^{2}+y^{2}=R^{2}$

Therefore, the cone has equation $z=\frac{Hr}{R}$ in polar coordinates.

This gives the above integral which evaluates to $\frac{1}{3}{\pi}R^{2}H$

12. Hello, Joanne!

a) Prove the surface area of a sphere of radius $a$ is: $4 \pi a^2$
We have the circle: . $x^2 + y^2\:=\:a^2\quad\Rightarrow\quad y \:=\:\sqrt{a^2-x^2}$ . (upper semicircle)

Area of a surface of revolution: . $A \;=\;2\pi\int^b_a y\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$

We have: . $y \:=\:\left(a^2-x^2\right)^{\frac{1}{2}}\quad\Rightarrow\quad\frac {dy}{dx} \:=\:\frac{-x}{\sqrt{a^2-x^2}}$

. . Then: . $\sqrt{1 + \left(\frac{dy}{dx}\right)} \:=\:\sqrt{1 + \left(\frac{-x}{\sqrt{a^2-x^2}}\right)^2} \:=\:\sqrt{1 + \frac{x^2}{a^2-x^2}} \:=\:\sqrt{\frac{a^2}{a^2-x^2}}\:=\:\frac{a}{\sqrt{a^2-x^2}}$

So we have: . $A \;=\;2 \times 2\pi\int^a_0\sqrt{a^2-x^2}\cdot\frac{a}{\sqrt{a^2-x^2}}\,dx \;=\;4\pi a\int^a_0\,dx \;=\;4\pi a\bigg[x\bigg]^a_0$

. . $A \;=\;4\pi a[a - 0]\quad\Rightarrow\quad\boxed{A \;=\;4\pi a^2}$

b) Prove the area of a disk of radius $a$ is: $\pi a^2$
We have the circle: . $x^2 + y^2\:=\:a^2\quad\Rightarrow\quad y \:=\:\sqrt{a^2-x^2}$ . (upper semicircle)

The area under the semicircle is: . $A \;=\;\int^a_{\text{-}a}y\,dx$

The area of the entire circle is: . $A \;=\;2 \times \int^a_{\text{-}a}y\,dx \;=\;4 \int^a_0y\,dx$

We have: . $A \;=\;4\int^a_o\sqrt{a^2-x^2}\,dx$

Let: $x = a\sin\theta\quad\Rightarrow\quad dx = a\cos\theta\,d\theta$

Substitute: . $A \;=\;4\int a\cos\theta(a\cos\theta\,d\theta) \;=\;4a^2\int\cos^2\theta\,d\theta \;=\;2a^2\int(1 + \cos2\theta)d\theta$

. . $= \;2a^2\left(\theta + \frac{1}{2}\sin2\theta\right) \;=\;2a^2\left(\theta + \frac{1}{2}\cdot2\sin\theta\cos\theta\right) \;=\;2a^2(\theta + \sin\theta\cos\theta)$

Back-substitute: . $A \;=\;2a^2\left[\sin^{-1}\left(\frac{x}{a}\right) + \frac{x\sqrt{a^2-x^2}}{a^2}\right]^a_0$

Evaluate: . $A \;=\;2a^2\left[\left(\sin^{-1}(1) + 0\right) - \left(\sin^{-1}(0) + 0\right)\right]$

Therefore: . $A \;=\;2a^2\left[\frac{\pi}{2} + 0 - 0 - 0\right] \quad\Rightarrow\quad\boxed{A\;=\;\pi a^2}$

13. They're are many other ways to tackle these. Soroban just showed a few more.

14. why don't you use washers volume method. i remember a website, wolfram or something may have these

15. Look at the earlier posts. That's what was used.

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