Lecture 19: Triple Integrals with Cyclindrical Coordinates and

Spherical Coordinates, Double Integrals for Surface Area, Vector

Fields, and Line Integrals

Part I: Triple Integrals with Cylindrical and Spherical

Coordinates

Changing variables for Triple Integrals• Similar to changing variables for double integrals.• Key question: If the new variables are u,v,w, what

is the volume corresponding to • The change from to causes a change of in • The change from to causes a change of in • The change from to causes a change of in

Changing variables for Triple Integrals

• The box from to , to , and to corresponds to a parallelepiped of volume

Triple Integrals with Cylindrical Coordinates

• Similar to double integrals with polar coordinates.• Recall: , ,

Example: Volume of a cone

Triple Integrals with Spherical Coordinates• Recall: , ,

(we are expanding it out by the third column)

Example: Volume of a Sphere

Part II: Double Integrals for Surface Area

Surface Area• What is the surface area of the section

corresponding to dxdy?• This is roughly a parallelogram with sides and

Surface Area

• Example: Surface area of a sphere

• Taking ,

Part III: Vector Fields and Line Integrals

Objectives

• Be familiar with vector fields and how to compute line integrals.

• Know how to compute the work done on an object by a given force.

Corresponding Section of Simmons: 21.1

Vector Fields

• A vector field assigns a vector to every point in space.

• Example: • Example: • Vector fields can be sketched by drawing the

vector at each point.

Line Integrals

• Breaking up a path into many small segments , • How do we break our path into many small

segments? Choose a parametrization of the curve and integrate accordingly.

• Fact: All parametrizations of a curve give the same answer.

Line Integrals with Parametric Equations

• If we have parametric equations and a vector field , the line integral is

• Example: If , and then the line integral from to is

Kinetic Energy and Acceleration• The kinetic energy of an object is • If an object starts with velocity , break up into a

component in the object’s direction of motion and a component perpendicular to the object’s direct of motion. .

• Initially, and .

• In time dt,

Work• This says that the amount of work done on an

object is • Example: If we apply a constant force of 10N

to push a box 5m, then we have done 50J of work.

• Example: If we swing an object around in a circle, the force on the object is towards the center and perpendicular to the object’s motion, so no work is done.

Examples

• If we apply a force of on an object and move the object from to , the total work we do is

• To lift an object of mass m to height h, we must apply a upward force of for that distance. The total work we need to do is