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Math Help - areas, surfaces and volumes...

  1. #1
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    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

    .................................................. ...........
    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

    ...........................................
    how can i do the rest please. and what integration methods should I be using for each? thnx xxxx
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  2. #2
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    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}
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  3. #3
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    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?

    .........

    any suggestions for the other ones please?

    xx
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  4. #4
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    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?.
    Last edited by galactus; August 7th 2007 at 05:33 PM.
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  5. #5
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    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<br />

    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?
    for example,

    Quote Originally Posted by galactus View Post
    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?

    .............................

    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
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  6. #6
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    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.
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  7. #7
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    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
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  8. #8
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    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.
    .......................

    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
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  9. #9
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    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.
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  10. #10
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    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?
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  11. #11
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    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
    Last edited by galactus; August 8th 2007 at 01:21 PM.
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  12. #12
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    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}

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  13. #13
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    They're are many other ways to tackle these. Soroban just showed a few more.
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  14. #14
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    why don't you use washers volume method. i remember a website, wolfram or something may have these
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  15. #15
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    Look at the earlier posts. That's what was used.
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