# Volume of solid bounded by paraboloid and plane.

1. Nov 17, 2009

### philnow

1. The problem statement, all variables and given/known data

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?

2. Nov 17, 2009

### HallsofIvy

Staff Emeritus
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
$$\int\int\int_{z= (1/4)(x^2+y^2)}^4 dzdydx[/itex] or just write it as the double integral $\int\int ((1/4)(x^2+y^2)- 4)dydx$. Now, the limits of integration in x and y: When you graphed it you probably saw that the paraboloid and plane intersect where z= 4 and $4(4)= 16= x^2+ y^2$ which, projected to the xy-plane is the circle $x^2+ y^2= 16$ and the entire figure is inside that cylinder. So the limits of integration for x and y are given by that circle. One way to cover that circle is to take x from -4 to 4 and, for each x, y from [tex]-\sqrt{16- x^2}$$
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
$$\sqrt{16- y^2}$$.

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.

Last edited: Nov 17, 2009
3. Nov 17, 2009

### 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?

4. Nov 17, 2009

### 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.

5. Nov 17, 2009

### HallsofIvy

Staff Emeritus
I believe I answered that in the second paragraph of my response.

6. Nov 17, 2009

### HallsofIvy

Staff Emeritus
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, $4y- (1/4)(x^2y+ (1/3)y^3)$ evaluated between
$-\sqrt{16- x^2}$ 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).

But yes, it will be much simpler in polar coordinates.

7. Nov 18, 2009

### 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

8. Nov 19, 2009

### philnow

Could I get a confirmation that this is indeed correct?