Volume of solid bounded by paraboloid and plane.

[philnow]

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?

 

[HallsofIvy]

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 V= \int\int\int dV= \int\int\int dzdydx 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 \int\int\int (f(x,y)- g(x,y))dydx

Here the upper boundary is just z= 4 and the lower boundary is z= (1/4)(x^2+ y^2), You could integrate
\int\int\int_{z= (1/4)(x^2+y^2)}^4 dzdydx[/itex]<br /> or just write it as the double integral \int\int ((1/4)(x^2+y^2)- 4)dydx.<br /> <br /> 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 4(4)= 16= x^2+ y^2 which, projected to the xy-plane is the circle x^2+ y^2= 16 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<br /> -\sqrt{16- x^2}<br /> to <br /> \sqrt{16- x^2}<br /> Or, because of symmetry, y from -4 to 4 and, for each y, x from <br /> -\sqrt{16- y^2}<br /> to <br /> \sqrt{16- y^2}.<br /> <br /> Perhaps simplest, because of the circular symmetry, is to use polar coordinates with r going from 0 to 4 and \theta from 0 to 2\pi. That would be the same as setting up the entire in cylindrical coordinates: the boundaries would have equations z= 4 and 4z= r^2 in cylindrical coordinates.

 

[philnow]

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?

 

[philnow]

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.

 

[HallsofIvy]

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.

 

[HallsofIvy]

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.
\int_{x=-1}^1\int{y= -\sqrt{16-x^2}}^\sqrt{16-x^2} (4- (1/4)(x^2+ y^2)dydx[/itex]<br /> <br /> The first integral will give, of course, 4y- (1/4)(x^2y+ (1/3)y^3) evaluated between <br /> -\sqrt{16- x^2} and \sqrt{16-x^2} or 8\sqrt{1-x^2}+ 2(x^2\sqrt{1- x^2}+ (1/3)(1- x^2)^{3/2} and that looks like a candidate for a trig substitution like x= sin(t).<br /> <br /> But yes, it will be much simpler in polar coordinates.

 

[philnow]

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

 

[philnow]

Could I get a confirmation that this is indeed correct?