Finding volume using polar coordinates

In summary, the conversation discusses finding the volume of a region inside a paraboloid surface, between two given values of z. The person has already found the limits of integration but is having trouble calculating the integral. One person suggests using a trick to solve the integral, while another suggests converting to polar coordinates. The conversation ends with a detailed explanation of how to solve the integral using polar coordinates. The final answer is 50π.
  • #1
babbagee
74
0
Ok the question is

find the volume of the region inside the surface z = x2 + y2 and between z = 0 and z = 10.

Ok i have already found the limits of integration but i am having a hard time calculating the integral.

The limits are [tex]-{\sqrt{10-x^2}[/tex] <= y <= [tex]{\sqrt{10-x^2}[/tex]
[tex]-{\sqrt{10}[/tex] <= x <= [tex]{\sqrt{10}[/tex]

I am integrating with respect to y first and then x.

The problem is after i integrate with respect to y i get a really hard integral which i tried soliving by trignometric substitution but i just made more complicated.

Can someone help me out please

thanks
 
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  • #2
Here's a trick if you didn't manage the trig-substitution:
Volume inside parabolic=Volume of cylinder minus volume under parabolic.
The easiest way to do the last volume, is to change into polar coordinates.
 
  • #3
I think they are asking for the volume under the paraboloid because i integrated once and then i entered the rest of it into the calculator and got the same answer as the back of the book. So i don't think they are asking for the volume above the paraboloid even though it sounds like it. That is what i thought at first. I don't think i need to convert into polor coordinates becuase the book has not gone into that much detail about converting into different coordinate systems, that is the next chapter. My limits are based on the volume under the paraboloid. I was at first thing of taking the volume under the paraboloid and subtracting it from the the cube that the paraboloid is enclosed in, its the same concept as the cylinder. But i don't think this problem wants me to do all that. Is there any way i can calculate the second integral. and by the way the answer in the back of the book as 50pi, and my calculater gave an answer of 157.0796 which is 50pi.

Thanks
 
  • #4
Glad you found out.
I don't think the trig-sub is really that difficult..
(Haven't checked, though)
 
  • #5
ok, this is old but still a good example of cylindrical coordinates.
The rectangular integral to solve this would be

[tex]\int_{-\sqrt{10}}^{\sqrt{10}} \int_{-\sqrt{10-x^2}}^{\sqrt{10-x^2}} (x^2+y^2) dydx}[/tex]

which if you plug into a computer solver is no problem, but unfortunately you will need another method to solve this integral by hand.
First off, why do you convert to polar coordinates here? Well if you think about the projection(or a shadow) of the paraboloid
on the xy-plane then you can see a circle with radius of [tex]{\sqrt{10}}[/tex] and what better to represent circles than polar coordinate.
If we where dealing with a function in [tex]R^2[/tex] we know that [tex]x=rcos{\theta}; y=rsin{\theta}[/tex], which is true in [tex]R^3[/tex] and we just leave [tex]z=z[/tex].
[tex]dA[/tex] will become [tex]rdrd{\theta}[/tex] (this can be explained if you think of how polar coordinates work, infinitesimally small blocks of a circle can be
thought of as a normal rectangle with sides [tex]rdr[/tex] and [tex]d{\theta}[/tex] making the area [tex]rdrd{\theta})[/tex] so [tex]dA=dxdy=rdrd{\theta}[/tex] also notice that
[tex]x^2+y^2=r^2(cos^2{\theta}+sin^2{\theta})=r^2[/tex]. We know that [tex]r[/tex] bounds are [tex]0\underline{<}r\underline{<}{\sqrt{10}}[/tex] and [tex]0\underline{<}{\theta}\underline{<}2{\pi}[/tex] so the new integral to solve is
[tex]\int_{0}^{2{\pi}} \int_{0}^{\sqrt{10}} r^2 rdrd{\theta} = \int_{0}^{2{\pi}} \int_{0}^{\sqrt{10}} r^3 drd{\theta} = \int_{0}^{2{\pi}} d{\theta} \int_{0}^{\sqrt{10}} r^3 dr = \frac{2{\pi}}{4}\sqrt{10^4} = \frac{200{\pi}}{4} = 50{\pi} [/tex]

I hope this helps anyone that may be having similar problems with this type of problem.
 

1. What is a double integral?

A double integral is a mathematical concept that involves finding the signed area between a surface and a specified region in a two-dimensional space. It is used to calculate volumes, surface areas, and other physical quantities in mathematics and physics.

2. How do you solve a double integral?

To solve a double integral, you first need to set up the limits of integration for both the inner and outer integrals. Then, evaluate the inner integral, treating the outer integral as a constant. Finally, evaluate the outer integral using the result from the inner integral.

3. What are the different methods for solving a double integral?

There are several methods for solving a double integral, including the rectangular method, the polar method, and the change of variables method. The method used will depend on the shape of the region and the function being integrated.

4. What are the applications of double integrals?

Double integrals have various applications in mathematics and physics. They are commonly used to calculate volumes, surface areas, and moments of inertia. They also have applications in calculating probabilities, electric fields, and fluid dynamics.

5. What are some common mistakes to avoid when solving a double integral?

Some common mistakes to avoid when solving a double integral include incorrect setup of the limits of integration, forgetting to apply the chain rule when using the change of variables method, and forgetting to multiply by the appropriate constants when converting between rectangular and polar coordinates.

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