Find Volume of Object Bounded by z=4-4(x^2+y^2) and z=(x^2+y^2)^2 -1

  • Thread starter Mindscrape
  • Start date
  • Tags
    Volume
In summary, the problem involves finding the volume of an object bounded by two equations in terms of z, z=4-4(x^2+y^2) for the top and z=(x^2+y^2)^2 -1 for the bottom. After converting to polar coordinates and integrating in the first quadrant, the resulting volume is 7pi/2. However, the correct answer is 8pi/3 as there was a mistake in setting up the integration. In terms of x and y, the region of integration can be visualized as a plastic Easter egg, making the integration more complicated.
  • #1
Mindscrape
1,861
1
The top of an object is bounded by [tex] z=4-4(x^2+y^2)[/tex] and the bottom is bounded by [tex]z=(x^2+y^2)^2 -1 [/tex], find the volume.

So I converted to polar coordinates such that
top: [tex]z=4-4r^2[/tex]
bot: [tex]z=r^2 - 1[/tex]

Then I integrated the first quadrant and multiplied by 4, with the bounds of dzdrdø such that the volume was:
[tex]V= 4[ \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=0}^{z=4-4r^2} r dz dr d \theta + \int_{\theta=0}^{\pi/2} \int_{r=0}^{r=1} \int_{z=r^2-1}^{z=0} r dz dr d\theta ][/tex]

The answer I got was 7pi/2, but the answer from the book is 8pi/3. I think I have done all the arithematic right, but did I set it up right?
 
Last edited:
Physics news on Phys.org
  • #2
what is the region of integration in terms of x&y?
 
  • #3
It isn't given, the problem just gave the object in terms of its top and bottom. I sketched it out, and the object is essentially one of those plastic easter eggs (where the top is longer and narrower than the bottom).

But, just going back then in terms of x&y the integration will be the very sticky:
[tex]V= 4[ \int_{y=0}^{y=1} \int_{x=0}^{x=\sqrt{1-y^2}} \int_{z=0}^{z=4-4(x^2+y^2)} dz dx dy + \int_{y=0}^{y=1} \int_{x=0}^{x=\sqrt{1-y^2}} \int_{z=(x^2+y^2)^2-1}^{z=0}dz dx dy][/tex]
Which does not look appealing to calculate.
 
Last edited:
  • #4
Ah, but [tex]z=(x^2+y^2)^2 -1 [/tex] transforms into [tex]z=r^4 -1 [/tex]! power 4 not 2.
 
  • #5
Oh no, you are right! Oops.

Thanks.
 

Related to Find Volume of Object Bounded by z=4-4(x^2+y^2) and z=(x^2+y^2)^2 -1

1. What is the equation for the given surface?

The given surface is bounded by the equations z=4-4(x^2+y^2) and z=(x^2+y^2)^2 -1.

2. How do you find the volume of an object bounded by two surfaces?

To find the volume of an object bounded by two surfaces, you can use the double integral method. First, find the limits of integration by setting the two equations equal to each other and solving for x and y. Then, set up the double integral with the limits of integration and integrate the function (1) with respect to x and y.

3. What is the significance of the equations used to bound the object?

The equations used to bound the object represent the surfaces that form the boundaries of the object. These equations help define the shape and size of the object and can be used to calculate its volume.

4. What is the relationship between the two equations used to bound the object?

The two equations are related because they both involve the variables x and y. The first equation, z=4-4(x^2+y^2), represents a paraboloid that opens downward, while the second equation, z=(x^2+y^2)^2 -1, represents a paraboloid that opens upward. Together, these two surfaces form the boundaries of the object.

5. Can the volume of the bounded object be negative?

No, the volume of the bounded object cannot be negative. Volume is a measure of the space occupied by an object, and it is always a positive quantity.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
690
  • Calculus and Beyond Homework Help
Replies
2
Views
595
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Calculus and Beyond Homework Help
Replies
14
Views
824
  • Calculus and Beyond Homework Help
Replies
11
Views
917
  • Calculus and Beyond Homework Help
Replies
3
Views
512
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
516
Back
Top