TRIPLE INTEGRALS MATH 200 GOALS Be able to set up and evaluate - PowerPoint PPT Presentation

MATH 200 WEEK 9 - WEDNESDAY TRIPLE INTEGRALS

MATH 200 GOALS ▸ Be able to set up and evaluate triple integrals using rectangular, cylindrical, and spherical coordinates

MATH 200 TRIPLE INTEGRALS ▸ We integrate functions of three variables over three dimensional solids ��� dV F ( x, y, z ) dV S S ▸ Chop the solid S up into a (x 0 ,y 0 ,z 0 ) bunch of cubes with volume dV ▸ Pick a point in each cube and evaluate F there ▸ Add up all of these products (F•dV)

MATH 200 INTERPRETATIONS ▸ If we think of F(x,y,z) as giving the density of the solid S at (x,y,z), then the triple integral gives us the dV mass of S S ▸ If F(x,y,z) = 1, then the (x 0 ,y 0 ,z 0 ) integral gives us the volume of S

MATH 200 EXAMPLE 1 � 1 / 3 � 1 � 1 / 3 z =1 � π � π � 1 � 2 z 2 x sin( xy ) zx sin( xy ) dzdydx = dydx � � 0 0 0 0 0 z =0 � 1 / 3 � π 1 = 2 x sin( xy ) dydx 0 0 � 1 / 3 y = π �� � 1 − 1 � = x cos( xy ) 2 x dx � � 0 y =0 � 1 / 3 y = π � − 1 � = 2 cos( xy ) dx � � 0 y =0 � 1 / 3 − 1 2 cos( π x ) + 1 = 2 dx 0 1 / 3 � = − 1 2 π sin( π x ) + 1 d � dy cos( xy ) = − x sin( xy ) 2 x � � 0 sin( xy ) dy = − 1 � √ x cos( xy ) + C 4 π + 1 3 = − 6

MATH 200 LOTS OF WAYS TO SETUP ▸ Let’s set up a few triple integrals for the volume of the solid bounded by y 2 + z 2 = 1 and y = x in the first octant ▸ This means, we’ll just integrate F(x,y,z) = 1 ▸ Here’s what the solid looks like: ▸ A sketch will really help with these problems

MATH 200 ▸ Let’s say we want to integrate in the order dzdydx ▸ Once we integrate with respect to z, z is gone ▸ Visually, we can think of flattening the solid onto the xy-plane ▸ The top bound for z is the surface z 2 =(1-y 2 ) 1/2 ▸ The bottom bound is the xy-plane, z 1 =0

MATH 200 ▸ Once we’ve flattened out in the z-direction, we have a double integral to set up, which we already know how to do! ▸ We have y 1 =x & y 2 =1 and x 1 =0 & x 2 =1 ▸ So the triple integral becomes � √ � 1 � 1 1 − y 2 1 dzdydx 0 0 x

MATH 200 ▸ Alternatively, we could have gone with dzdxdy ▸ In this case all that changes is the outer double integral ▸ Going back to the flattened image on the xy-plane, we get x 1 =0 & x 2 =y and y 1 =0 & y 2 =1 � √ � 1 � y 1 − y 2 1 dzdxdy 0 0 0

MATH 200 ▸ We could also not start with z. For example, let’s try dxdzdy ▸ Integrating with respect to x first will flatten the picture onto the yz-plane ▸ On “top” (meaning further out towards us), we have x 2 =y ▸ On the “bottom” (meaning yz-plane further back) we have x 1 =0

MATH 200 ▸ Now we just set up the bounds for the outer two integrals based on the flattened image on the yz- plane ▸ z 1 =0 & z 2 =(1-y 2 ) 1/2 ▸ y 1 =0 & y 2 =1 ▸ So the triple integral becomes � √ � 1 � y 1 − y 2 1 dxdzdy 0 0 0

MATH 200 ▸ Pick one of these three to integrate: � 1 � 1 � √ � 1 � √ � y 1 − y 2 1 − y 2 1 dzdydx 1 dzdxdy 0 0 0 0 0 x � 1 � √ � y 1 − y 2 1 dxdzdy 0 0 0 ▸ With the dzdydx integral, we end up needing trig substitution to perform the second integration, so we should go with the second or third option

MATH 200 √ � √ � y � y � 1 � 1 1 − y 2 1 − y 2 � � 1 dzdxdy = z dxdy � � 0 0 0 0 0 0 � y � 1 1 − y 2 dxdy � = 0 0 � 1 y � � � 1 − y 2 x dy = � � 0 0 � 1 � 1 − y 2 dy y = 0 � 0 − 1 √ u du = 2 1 0 �� � 2 = − 1 3 u 3 / 2 � � 2 � 1 = 1 3

MATH 200 EXAMPLE 2 ��� z √ y dV where G is the solid enclosed by z = y , y = x 2 , y = 4, Evaluate G and z = 0.

MATH 200 ▸ Let’s try dzdydx first ▸ z 1 =0 and z 2 =y ▸ Flatten the solid onto the xy- plane ▸ Now for y we have… ▸ y 1 =x 2 & y 2 =4 ▸ Finally, ▸ x 1 =-2 & x 2 =2

MATH 200 � y � y � 2 � 4 � 2 � 4 z √ y dzdydx = 2 z √ y dzdydx x 2 x 2 − 2 0 0 0 � 2 � 4 y � 1 2 z 2 √ y � = 2 dydx � WE HAVE TO BE CAREFUL � x 2 0 0 � 2 � 4 WHEN USING SYMMETRY: x 2 y 2 √ y dydx = 0 � 2 � 4 IT WORKS HERE BECAUSE x 2 y 5 / 2 dydx = BOTH THE FUNCTION WE’RE 0 � 2 INTEGRATING AND THE REGION 4 � 2 7 y 7 / 2 � = x 2 dx OVER WHICH WE’RE � � 0 INTEGRATING ARE SYMMETRIC � 2 4 � = 2 128 − x 7 � x 2 dx OVER THE PLANE X=0. � 7 � 0 2 �� = 2 � 128 x − 1 8 x 8 � � 7 � 0 = 2 7 (256 − 32) = 64

MATH 200 ▸ Let’s look at some alternative setups ▸ We could have started with x instead and done dxdzdy ▸ If we draw a line through the solid in the x-direction, it first hits the back half of the parabolic surface and then the front half of the parabolic surface ▸ If we then collapse the picture onto the yz-plane, we get this… � 4 � √ y � y z √ y dxdzdy −√ y 0 0

MATH 200 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES ▸ Cylindrical coordinates ▸ We already know from polar that dA = rdrd θ ▸ So, for dV we get rdrd θ dz ▸ Just replace dxdy or dydx with rdrd θ ▸ Spherical coordinates ▸ For now, let’s just accept that in spherical coordinates, dV becomes ρ 2 sin φ d ρ d φ d θ ▸ We’ll come back to why this is the case in the next section

MATH 200 EXAMPLE � √ � 2 � √ 4 − x 2 4 − x 2 − y 2 4 − x 2 − y 2 x 2 + y 2 dzdydx ▸ Consider the integral − √ 0 0 ▸ First let’s get a sense of what the region/solid looks like ▸ z 1 = -(4 - x 2 - y 2 ) 1/2 and z 2 = (4 - x 2 - y 2 ) 1/2 ▸ Squaring both sides of either equation, we get a sphere of radius 2 centered at (0,0,0): x 2 + y 2 + z 2 = 4 ▸ So we’re going from the bottom half of the sphere to the top half ▸ y 1 = 0 & y 2 = (4-x 2 ) 1/2 and x 1 = 0 & x 2 = 2 ▸ On the xy-plane, we go from the line y=0 to the top half of a circle of radius 2, but only from x=0 to x=2.

MATH 200

MATH 200 ▸ Setup in cylindrical coordinates ▸ Since z is common to rectangular and cylindrical, let’s start with that � � 4 − x 2 − y 2 = 4 − r 2 z = − ⇒ z = − � � 4 − x 2 − y 2 = 4 − r 2 z = ⇒ z = ▸ Now we can look at what remains on the xy-plane and convert that to polar (recall: cylindrical = polar + z) ▸ y 1 = 0 & y 2 = (4-x 2 ) 1/2 and x 1 = 0 & x 2 = 2 ▸ On the xy-plane, we go from the line y=0 to the top half of a circle of radius 2, but only from x=0 to x=2. ▸ r goes from 0 to 2 and θ goes from 0 to π /2

MATH 200 θ 2 = π 2 � 4 − r 2 z 2 = r 2 = 2 θ 1 = 0 r 1 = 0 � 4 − r 2 z 1 = − � √ � √ � 2 � √ � 2 � π / 2 4 − x 2 4 − x 2 − y 2 4 − r 2 4 − x 2 − y 2 ( x 2 + y 2 ) dzdydx = 4 − r 2 r 2 rdzdrd θ √ − √ 0 0 0 0 −

MATH 200 ▸ For spherical, let’s start with ρ : ▸ The sphere of radius 2 is simply ρ =2 ▸ The region starts at the origin: ρ =0 ▸ Remember, φ measures the angle taken from the positive z-axis ▸ In order to cover the quarter-sphere, φ needs to go from 0 to π . ▸ We already know what θ does from cylindrical coordinates ▸ The integrand (the function we’re integrating) is a little more involved…

MATH 200 x 2 + y 2 = ( ρ sin φ cos θ ) 2 + ( ρ sin φ sin θ ) 2 = ρ 2 sin 2 φ cos 2 θ + ρ 2 sin 2 φ sin 2 θ = ρ 2 sin 2 φ (cos 2 θ + sin 2 θ ) = ρ 2 sin 2 φ ▸ Lastly, we can’t forget about the “extra term,” ρ 2 sin φ : � √ � 2 � √ � 2 4 − x 2 4 − x 2 − y 2 � π / 2 � π 4 − x 2 − y 2 ( x 2 + y 2 ) dzdydx = ρ 2 sin 2 φ ρ 2 sin φ d ρ d φ d θ − √ 0 0 0 0 0 � 2 � π / 2 � π ρ 4 sin 3 φ d ρ d φ d θ = 0 0 0