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Triple Integral Limits Help. Cylindrical Coordinates

  • Thread starter bob29
  • Start date
  • #1
18
0

Homework Statement


Find the volume of the solid bounded by the paraboloids z=x^2+y^2 and z=36-x^2-y^2.

Answer is:
[tex]
324\pi
\\[/tex]

Homework Equations


r^2=x^2+y^2
x=rcos0
y=rcos0

The Attempt at a Solution


36-x^2+y^2=x^2+y^2\\
36=2x^2+2y^2
18=x^2+y^2
r^2=18

[tex]
V=\int_{0}^{2\pi} \int_0^{3\sqrt{2}} \int_{r^2}^{36-r^2} \left (1) \right dz.rdr.d\theta
[/tex]

---------------------------------------------------------------

Homework Statement


Use cylindrical coordinates to find the volume of inside the region cut from the sphere x^2+y^2+z^2=36 by the cylinder x^2+y^2=4.
Ans is:
[tex]
\frac{4\pi}{3}(216-32^{3/2})
[/tex]

The Attempt at a Solution


Diagram is of a sphere where a cylinder is inside of it.
x^2+y^2+z^2=4
r^2+z^2=4
[tex]
z=\sqrt{4-r^2}\\
[/tex]

x^2+y^2=4
r^2=2^2
r=2

[tex]
V=\int_{0}^{2\pi} \int_0^2 \int_{-\sqrt{36-r^2}}^{\sqrt{4-r^2}} \left (1) \right dz.rdr.d\theta
[/tex]

Studying for an exam and would appreciate the help to answering these questions that I am struggling on.
 
Last edited:

Answers and Replies

  • #2
360
0
The first problem looks good: like you I got [itex] 324 \pi [/itex].

Your setup for the second volume is almost correct, but not quite. Your upper limit of integration for z is incorrect. When you're picturing the region, remember that it's the sphere that gives you the upper and lower boundaries for the z coordinate.
 
  • #3
18
0
Isn't the region the cylinder? so it enters the bottom and exits the top of the cylinder. Sort of confused about which function determines the boundaries.

Thanks for confirming the 1st problem.
Guess I made a mistake in my integration of r where r^4/4 should have been r^4/2.
 

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