Contiune on 16.7 Triple Integrals - University of Notre Dame

1 Contiune on Triple Integrals Figure 1: ZZZ Z Z "Z u2 (x,y). #. f (x, y, z)dV = f (x, y, z)dz dA. E D u1 (x,y). Applications of Triple Integrals Let E be a solid region with a density function (x, y, z). RRR. Volume: V (E) = E. 1dV. RRR. Mass: m = E. (x, y, z)dV. Moments about the coordinate planes: ZZZ. Mxy = z (x, y, z)dV. E. ZZZ. Mxz = y (x, y, z)dV. E. ZZZ. Myz = x (x, y, z)dV. E. Center of mass: (x , y , z ). x = Myz /m , y = Mxz /m , z = Mxy /m . Remark: The center of mass is just the weighted average of the coordinate functions over the solid region. If (x, y, z) = 1, the mass of the solid equals its volume and the center of mass is also called the centroid of the solid.

2 Example Find the volume of the solid region E between y = 4 x2 z 2 and y = x2 + z 2 . 1. Soln: E is described by x2 + z 2 y 4 x2 z 2 over a disk D in the xz-plane whose radius is given by the intersection of the two surfaces: y = 4 x2 z 2 and y = x2 + z 2 . 4 x2 z 2 = x2 + z 2 x2 + z 2 = 2. So the radius is 2. Therefore ZZZ Z Z "Z 4 x2 z 2 # ZZ. V (E) = 1dV = 1dy dA = 4 2(x2 + z 2 )dA. E D x2 +z 2 D. 2 Z 2 2 2. 1. Z Z. = (4 2r 2)rdrd = 2r 2 r 4 = 4 . 0 0 0 2 0. Example Find the mass of the solid region bounded by the sheet z = 1 x2 and the planes z = 0, y = 1, y = 1 with a density function (x, y, z) = z(y + 2).

3 Figure 2: Soln: The top surface of the solid is z = 1 x2 and the bottom surface is z = 0 over the region D in the xy-plane which is bounded by the other equations in the xy-plane and the intersection of the top and bottom surfaces. The intersection gives 1 x2 = 0 x = 1. Therefore D is a square [ 1, 1] [ 1, 1]. ZZZ ZZZ Z Z "Z 2. #. 1 x m= (x, y, z)dV = z(y + 2)dV = z(y + 2)dz dA. E E D 0. 1 1 1 x2 1 1. 1. Z Z Z Z Z. = z(y + 2)dzdxdy = (1 x2 )2 (y + 2)dxdy =. 1 1 0 2 1 1. Z 1. 8. (y + 2)dy = 32/15. 15 1. Example Find the centroid of the solid above the paraboloid z = x2 + y 2 and below the plane z = 4.

4 Soln: The top surface of the solid is z = 4 and the bottom surface is z = x2 + y 2 over the region D defined in the xy-plane by the intersection of the top and bottom surfaces. 2. Figure 3: The intersection gives 4 = x2 + y 2 . Therefore D is a disk of radius 2. By the symmetry principle, x = y = 0. We only compute z : ZZZ Z Z Z 4 ZZ Z 2 Z 2. 2 2. m= 1dV = 1dz dA = 4 (x + y )dA = (4 r 2 )rdrd = 8 . E D x2 +y 2 D 0 0. Z Z Z 4 . 1. ZZZ ZZ. Mxy = zdV = zdz dA = 8 (x2 + y 2)2 dA =. E D x2 +y 2 D 2. 2 2 2 . 1 1 62. Z Z Z. (8 r 4 )rdrd = [4r 2 r ] d = 64 /3. 0 0 2 0 12 0. Therefore z = Mxy /m = 8/3 and the centroid is (0, 0, 8/3).

5 Triple Integrals in Cylindrical and Spherical Coordinates 1. Triple Integrals in Cylindrical Coordinates A point in space can be located by using polar coordinates r, in the xy-plane and z in the vertical direction. Some equations in cylindrical coordinates (plug in x = r cos( ), y = r sin( )): Cylinder: x2 + y 2 = a2 r 2 = a2 r = a;. Sphere: x2 + y 2 + z 2 = a2 r 2 + z 2 = a2 ;. Cone: z 2 = a2 (x2 + y 2) z = ar;. Paraboloid: z = a(x2 + y 2) z = ar 2 . The formula for Triple integration in cylindrical coordinates: If a solid E is the region between z = u2 (x, y) and z = u1 (x, y) over a domain D in the xy-plane, which is described in polar coordinates by , h1 ( ) r h2 ( ), we plug 3.

6 Figure 4: in x = r cos( ), y = r sin( ). ZZZ Z Z "Z u2 (x,y). #. f (x, y, z)dV = f (x, y, z)dz dA =. E D u1 (x,y). Z Z h2 ( ) Z u2 (r cos ,r sin ). f (r cos , r sin , z)rdzdrd . h1 ( ) u1 (r cos ,r sin ). Note: dV rdzdrd . zdV where E is the portion of the solid sphere x2 + y 2 + z 2 9. RRR. Example Evaluate E. that is inside the cylinder x2 + y 2 = 1 and above the cone x2 + y 2 = z 2 . Figure 5: p . Soln: The topp surface is z = u2 (x, y) = 9 x2 y 2 = 9 r 2 and the bottom surface is z = u1 (x, y) = x2 + y 2 = r over the region D defined by the intersection of the top (or 4. bottom) and the cylinder which is a disk x2 + y 2 1 or 0 r 1 in the xy-plane.

7 ZZZ Z Z "Z 9 r2 # Z 2 Z 1 Z 9 r2. zdV = zdz dA = zrdzdrd =. E D r 0 0 r 2 1 2 1 2 . 1 1. Z Z Z Z Z. [9 2r 2 ]rdrd = [9r 2r 3 ]drd = [9/4 1/4]d = 4 . 0 0 2 0 0 2 0. Example Find the volume of the portion of the sphere x2 +y 2 +z 2 = 4 inside the cylinder (y 1)2 + x2 = 1. Figure 6: p p Soln: The top surface is z = 4 x2 y2 = 4 r 2 and the bottom is z = 4 x2 y 2 =.. 4 r 2 over the region D defined by the cylinder equation in the xy-plane. So rewrite the cylinder equation x2 + (y 1)2 = 1 as x2 + y 2 2y + 1 = 1 r 2 = 2r sin( ) r = 2 sin( ).. ZZZ ZZ Z 4 r 2 Z Z 2 sin( ) Z 4 r 2. V (E) = 1dV =.

8 1dzdA = . 1rdzdrd =. E D 4 r 2 0 0 4 r 2. Z Z 2 sin( ) . 2r 4 r 2 drd (by substitution u = 4 r2 ) =. 0 0.. 2. Z. [(4 4 sin2 ( ))3/2 (4)3/2 ]d (use identity 1 = cos2 ( ) + sin2 ( )) =. 0 3. Z /2 Z . 16 16 16. Z. 3 3. [1 | cos( )| ]d = [1 cos ( )]d + [1 + cos3 ( )]d =. 0 3 0 3 /2 3. Z /2 Z . 16 16. [1 (1 sin2 ) cos ]d + [1 + (1 sin2 ) cos ]d =. 0 3 /2 3. /2. 16/3[( sin + sin3 /3)|0 + ( + sin sin3 /3)| /2 ] = 16 /3 64/9. 2. Triple Integrals in Spherical Coordinates 5. Figure 7: In spherical coordinates, a point is located in space by longitude, latitude, and radial distance. Longitude: 0 2.

9 Latitude: 0 ; p Radial distance: = x2 + y 2 + z 2 . From r = sin( ). x = r cos( ) = sin( ) cos( ). y = r sin( ) = sin( ) sin( ). z = cos( ). Some equations in spherical coordinates: Sphere: x2 + y 2 + z 2 = a2 = a Cone: z 2 = a2 (x2 + y 2) cos2 ( ) = a2 sin2 ( ). Cylinder: x2 + y 2 = a2 r = a or sin( ) = a Figure 8: Spherical wedge element 6. The volume element in spherical coordinates is a spherical wedge with sides d , d , rd . Replacing r with sin( ) gives: dV = 2 sin( )d d d . For our Integrals we are going to restrict E down to a spherical wedge. This will mean a b, , c d, ZZZ Z Z d Z b f (x, y, z)dV = f ( sin( ) cos( ), sin( ) sin( ), cos( )) 2 sin( )d d d.

10 E c a Figure 9: One example of the sphere wedge, the lower limit for both and are 0. The more general formula for Triple integration in spherical coordinates: If a solid E is the region between g1 ( , ) g2 ( , ), , c d, then ZZZ Z Z d Z g2 ( , ). f (x, y, z)dV = f ( sin( ) cos( ), sin( ) sin( ), cos( )) 2 sin( )d d d . E c g1 ( , ). Example Find the volume of the solid region above the cone z 2 = 3(x2 + y 2 ) (z 0). and below the sphere x2 + y 2 + z 2 = 4. Soln: The sphere x2 + y 2 + z 2 = 4 in spherical coordinates p is = 2. The cone z 2 =. 2 2 2 2. 3(x + y ) (z 0) in spherical coordinates is z = 3(x + y ) = 3r cos( ) =.