Volume of solid bounded by paraboloid and plane.

In summary, the volume of the solid bounded by the paraboloid and the plane z=4 is 4(4)= 16. The limits of integration for x and y are given by the circle x^2+ y^2= 16 and the entire figure is inside that cylinder. One way to cover that circle is to take x from -4 to 4 and, for each x, y from -\sqrt{16- x^2} to \sqrt{16- x^2}. or, because of symmetry, y from -4 to 4 and, for each y, x from -\sqrt{16- y^2} to [tex]\sqrt{16
  • #1
philnow
83
0

Homework Statement



Hi. I'm asked to find the volume of the solid bounded by the paraboloid

4z=x^2 + y^2 and the plane z=4

I have drawn the graph in 3D but I'm unsure of how to set up the integral. Also, how does one decide to use double integrals/triple integrals when finding volume?
 
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  • #2
philnow said:

Homework Statement



Hi. I'm asked to find the volume of the solid bounded by the paraboloid

4z=x^2 + y^2 and the plane z=4

I have drawn the graph in 3D but I'm unsure of how to set up the integral. Also, how does one decide to use double integrals/triple integrals when finding volume?
You could use both! But since [itex]V= \int\int\int dV= \int\int\int dzdydx[/itex] but since the integral of "dz" is just z, if the boundaries can be written as z= f(x,y) and z= g(x,y), then that triple integral just reduces to the double integral [itex]\int\int\int (f(x,y)- g(x,y))dydx[/itex]

Here the upper boundary is just z= 4 and the lower boundary is [itex]z= (1/4)(x^2+ y^2)[/itex], You could integrate
[tex]\int\int\int_{z= (1/4)(x^2+y^2)}^4 dzdydx[/itex]
or just write it as the double integral [itex]\int\int ((1/4)(x^2+y^2)- 4)dydx[/itex].

Now, the limits of integration in x and y: When you graphed it you probably saw that the paraboloid and plane intersect where z= 4 and [itex]4(4)= 16= x^2+ y^2[/itex] which, projected to the xy-plane is the circle [itex]x^2+ y^2= 16[/itex] and the entire figure is inside that cylinder. So the limits of integration for x and y are given by that circle. One way to cover that circle is to take x from -4 to 4 and, for each x, y from
[tex]-\sqrt{16- x^2}[/tex]
to
[tex]\sqrt{16- x^2}[/tex]
Or, because of symmetry, y from -4 to 4 and, for each y, x from
[tex]-\sqrt{16- y^2}[/tex]
to
[tex]\sqrt{16- y^2}[/tex].

Perhaps simplest, because of the circular symmetry, is to use polar coordinates with r going from 0 to 4 and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex]. That would be the same as setting up the entire in cylindrical coordinates: the boundaries would have equations z= 4 and [itex]4z= r^2[/itex] in cylindrical coordinates.
 
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  • #3
Thanks for the reply. If I wanted to set it up using polar coordinates (I haven't covered cylindrical coordinates yet) I would take r between 0 and 4, theta between 0 and 2pi, but over what function would I integrate?
 
  • #4
I can't seem to evaluate the integral using the "dydx" method, so I'm thinking this problem needs to be done using polar coordinates.
 
  • #5
philnow said:
Thanks for the reply. If I wanted to set it up using polar coordinates (I haven't covered cylindrical coordinates yet) I would take r between 0 and 4, theta between 0 and 2pi, but over what function would I integrate?
I believe I answered that in the second paragraph of my response.
 
  • #6
philnow said:
I can't seem to evaluate the integral using the "dydx" method, so I'm thinking this problem needs to be done using polar coordinates.
Sure you can.
[tex]\int_{x=-1}^1\int{y= -\sqrt{16-x^2}}^\sqrt{16-x^2} (4- (1/4)(x^2+ y^2)dydx[/itex]

The first integral will give, of course, [itex]4y- (1/4)(x^2y+ (1/3)y^3)[/itex] evaluated between
[itex]-\sqrt{16- x^2}[/itex] and [itex]\sqrt{16-x^2}[/itex] or [itex]8\sqrt{1-x^2}+ 2(x^2\sqrt{1- x^2}+ (1/3)(1- x^2)^{3/2}[/itex] and that looks like a candidate for a trig substitution like x= sin(t).

But yes, it will be much simpler in polar coordinates.
 
  • #7
Well, I'm not sure what part of your post corresponds to the second paragraph >.<

So does this integral look correct?

Double integral from 0 to 4 and from 0 to 2pi

(4-(x^2+y^2)/4)*r d(theta)dr = (4 - r/4)*r d(theta)dr
 
  • #8
Could I get a confirmation that this is indeed correct?
 

1. What is the formula for finding the volume of a solid bounded by a paraboloid and a plane?

The formula for finding the volume of a solid bounded by a paraboloid and a plane is V = ∫∫∫ dV = ∫∫∫ dxdydz, where the limits of integration are determined by the intersection points of the paraboloid and plane.

2. How do you determine the limits of integration for finding the volume of this solid?

The limits of integration for finding the volume of a solid bounded by a paraboloid and a plane can be determined by setting the equations of the paraboloid and plane equal to each other and solving for the intersection points. These points will then determine the limits for each variable in the triple integral.

3. Can you use any other method besides triple integration to find the volume of this solid?

Yes, the volume of a solid bounded by a paraboloid and a plane can also be found using the disk or shell method, depending on the orientation of the paraboloid and plane. These methods involve integrating over a single variable and can be easier to use in certain situations.

4. How do you determine which method to use for finding the volume of this solid?

The method used to find the volume of a solid bounded by a paraboloid and a plane depends on the orientation of the solid and the ease of integration. The disk method is typically used when the solid is oriented with the axis of revolution being the same as the axis of integration, while the shell method is used when the solid is oriented with the axis of revolution perpendicular to the axis of integration.

5. Can you generalize the formula for finding the volume of a solid bounded by a paraboloid and a plane?

Yes, the formula for finding the volume of a solid bounded by a paraboloid and a plane can be generalized as V = ∫∫∫ f(x,y) dV = ∫∫∫ f(x,y) dxdydz, where f(x,y) represents the function defining the paraboloid and plane and the bounds of integration are determined by the intersection points of the two surfaces.

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