Double integral polar coordinates trouble please help!

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Homework Statement



Consider the volume of a solid bounded by the cone: z = sqrt(x^2 + y^2) and the top half of the sphere x^2 + y^2 + z^2 = 18 that is for z >= 0

Using cylindrical coordinates, express the volume as a double integral.


Homework Equations


easy to sketch.. we can pretty easily figure r = 3 and theta = 2 pi


The Attempt at a Solution



I got integral of sqrt(18 - r^2) r.dr.d(theta) over r = { (rcos(theta), rsin(theta) | 0 <= r <= 3, 0 <= theta <= 2pi


but answer says... [sqrt(18-r^2) - r ] r.dr.d(theta) where have i gone wrong


thanks in advance for the help!
 

Answers and Replies

  • #2
Hootenanny
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Remember that we are dealing with cylindrical coordinates here, *not* spherical coordinates. So r2 = x2 + y2 and the volume element is r.dr.d[itex]\theta[/itex] dz.

So let's try again. Can you express the equation of the cone in terms of *cylindrical* coordinates?
 
  • #3
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im so lost :( where did the other r come from?
 
  • #4
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also r.dr.d(theta).dz is triple integral not double?
 
  • #5
Hootenanny
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im so lost :( where did the other r come from?
Which r? The expression, [itex]r dr d\theta dz[/itex] is simply the volume element in cylindrical coordinates. In Cartesian coordinates, the equivalent expression would be [itex] dx dy dz[/itex].
also r.dr.d(theta).dz is triple integral not double?
Indeed it is. Volume integrals *must* be triple integrals.
Consider the volume of a solid bounded by the cone: z = sqrt(x^2 + y^2) and the top half of the sphere x^2 + y^2 + z^2 = 18 that is for z >= 0

Using cylindrical coordinates, express the volume as a double integral.
However, triple integrals may be expressed as a double integral by evaluating one of the integrals, in this case the integral with respect to z.
 
  • #6
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[sqrt(18-r^2) - r ] i mean the r bold and underlined.

So i went for z limits o to sqrt(18)

with R = {(x,y,z) in polar form | 0 <= z <= sqrt(18), 0 <=r <= 3, 0 <= theta <= 2pi}

for what integral??? this is so confusing.. its just a Q for a practise exam is there anyway you could work it through abit? I find i understand much better if i can look at it to see how it works.. thankyou
 
  • #7
Hootenanny
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[sqrt(18-r^2) - r ] i mean the r bold and underlined.
We'll get to that eventually.
its just a Q for a practise exam is there anyway you could work it through abit? I find i understand much better if i can look at it to see how it works.. thankyou
That's not the way it works here. I'll guide you through the question, which will help you understand it better, but I won't provide a complete or partial solution.

Can you try answering my question:
So let's try again. Can you express the equation of the cone in terms of *cylindrical* coordinates?
In other words instead of writing

[tex]z = f\left(x,y\right)[/tex]

Write it in the form

[tex] z = g\left(r,\theta\right)[/tex]
 
  • #8
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so z=r for the cone in cylindrical form. now we are minusing the area from under the cone from the area under the half sphere right? so thats where we get our sqrt(18-r^2) - r from? because the z value ranges from z= sqrt(18-r^2) to z=r
 
  • #9
Hootenanny
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so z=r for the cone in cylindrical form.
Correct.
now we are minusing the area from under the cone from the area under the half sphere right? so thats where we get our sqrt(18-r^2) - r from? because the z value ranges from z= sqrt(18-r^2) to z=r
Let's take it one step at a time. So, as you correct write we also have the equation of the upper half sphere:

[tex]z = \sqrt{18-r^2}[/tex]

and for the cone:

[tex]z = r[/tex]

Now, we need to figure out the limits. Note that the limits will be different for the sphere and cone. What are the limits for the sphere?
 
  • #10
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Sorry had to sleep.. limits for sphere would be sqrt(18-r^2) and 0 and for the cone would be sqrt(18) and 0?
 
  • #11
Hootenanny
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Sorry had to sleep.. limits for sphere would be sqrt(18-r^2) and 0 and for the cone would be sqrt(18) and 0?
You need to be careful here, either the upper or lower limit in each case should be a function of the radius. Your limits for there sphere are correct: the smallest value z can take is zero (at the origin) and the largest value z can take is on the surface of the sphere (i.e. z = (18-r2)1/2.

Now can you apply the same logic to the cone? What is the smallest value that z can take inside the cone. What is the largest value?
 

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