Funny looking equation for a paraboloid.

In summary, the conversation discusses finding the volume between two paraboloids and suggests using a double integral to solve it. The bounds can be found by setting z=0 and solving for x and y. It is recommended to use polar coordinates for this problem.
  • #1
mishimakaz
4
0

Homework Statement


Find the volume between 2 paraboloids.


Homework Equations



z = 4x^2+8y^2
z= 30-x^2-y^2

The Attempt at a Solution



So I switched the variables to polar coordinates.
z = 30-r^2
z=4r^2*sin^2(θ)+8r^2*cos^2(θ)

Now I want to solve for r. However I get a very ugly looking value for r that does not integrate very well.
 
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  • #2
Why do you want to solve that for r? You want to do a double integral of zupper - zlower over the common xy domain. Figure out the equation of that domain and it might suggest a change of variables to you.
 
  • #3
I recommend two things:

1. No change of variables or coordinate systems, it just makes things more intricate.
2. Find the bounds first. They aren't as complex as you would imagine.

All of this assuming, you will use a double integral.

To find the upper paraboloid just evaluate the two at a point inside of the bounds you find.
 
  • #4
To find the bounds, I would set z=0 and then solve for x and then y. For example...y = sqrt(24-x^2).
 
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  • #5
Instead of solving for y, if you keep the x's and y's on the same side, you should see a familiar function in there.
 

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  • #6
m.w.lever said:
Instead of solving for y, if you keep the x's and y's on the same side, you should see a familiar function in there.

equation of a circle of sorts?
 
  • #7
yeah, I added an attachment, to my previous post, that shows the two paraboloids from the top. If you find the function for the boundary shape, you can find your bounds of x and y. Remember, you only need xmin xmax ymin and ymax
 
  • #8
m.w.lever said:
yeah, I added an attachment, to my previous post, that shows the two paraboloids from the top. If you find the function for the boundary shape, you can find your bounds of x and y. Remember, you only need xmin xmax ymin and ymax

No. You will have an ellipse shaped boundary which does not give constant limits. And a substitution similar to, but not exactly the same as, polar coordinates will be very helpful. You need to set the two z values equal to get the equation of the ellipse.
 
  • #9
LCKurtz is right, how could I be so silly, you can still get to the ellipse in cartesian, but for the bounds you will definitely need polar coordinates of some sort.
 

1. What is a paraboloid?

A paraboloid is a three-dimensional shape that resembles a bowl or a lens. It is formed by rotating a parabola around its axis.

2. Why does the equation for a paraboloid look funny?

The equation for a paraboloid, which is in the form of z = ax2 + by2, may look unusual because it contains two variables and a squared term. This is because a paraboloid is a curved shape that has a different slope in the x and y directions.

3. How is a paraboloid different from a parabola?

A parabola is a two-dimensional curve, while a paraboloid is a three-dimensional shape. A parabola has a single axis of symmetry, while a paraboloid has two axes of symmetry.

4. What are the common uses of paraboloids?

Paraboloids have many practical applications in science and engineering. They are used in satellite dishes, telescopes, reflectors, and solar cookers. They are also found in architecture, such as in domes and arches.

5. How is a paraboloid equation derived?

The equation for a paraboloid can be derived from the general equation of a quadratic surface, which is in the form of ax2 + by2 + cz2 + dxy + exz + fyz + gx + hy + iz + j = 0. By setting the parameters for c, d, e, f, g, h, i, and j to zero, the equation simplifies to z = ax2 + by2, which is the equation for a paraboloid.

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