Lecture 1 – Volumes Area – the entire 2-D region was sliced into strips Before width(x) was introduced, only dealing with length f(x) a b Volume – same concept, 3-D solid is sliced into strips Before width is introduced, only dealing with 2-D area

Solids of revolution Formed when revolving a region around a given line (axis). f(x) Revolve over x-axis x Follow one slice of the region (strip) as is gets swept out twice. 1st sweep: generates an area. 2nd sweep: generates a volume. Infinite number of disks used, hence Riemann sum turns to integral.

Example 1 Find the volume of the solid generated by revolving the region bounded the given curves around the x-axis. f(x) x 1 2

Example 2 Find the volume of the solid generated by revolving the region bounded the given curves around the x-axis. f(x) x 1 2 3

Washer method What happens when the region is revolved about a line but there is a gap between the two? Then the solid generated has a hole. f(x) Revolve over x-axis x

Example 3 Find the volume of the previous solid.

Example 4 Find the volume of the solid generated by revolving the region from Example 1 around the line y = -1. f(x) x 1 2

Lecture 2 – More Volumes Example 5 Find the volume of the solid generated by revolving the region bounded the given curves around the y-axis.

Example 6 Find the volume of the solid generated by revolving the region bounded the given curves around the x = 5. 5

Shells are created when strip is parallel to axis of revolution. What if you prefer to figure out everything in terms of the x-axis? How can volume work if revolution is around vertical axis? Disks(Washers) – when strip is perpendicular to axis of revolution Shells are created when strip is parallel to axis of revolution.

Example 7 Find the volume of the solid generated by revolving the region bounded the given curves around the y-axis using shells.

Example 8 Find the volume of the solid generated by revolving the region bounded the given curves around the x = 5 using shells. 5

Lecture 3 – Arc Length Arc Length – curve has a certain length Estimate the length of the given curve through appropriate choices for lower and upper bounds. 4 4 3 3 2 2 1 1

Lengths (in general) Use more and more hypotenuse lengths to better the approximations. Consider one segment out of n pieces and then let n go to infinity. f(x) x

Example 1 Find the length of the given curve. 4 3 2 1

Example 2 Find the length of the given curve. f(x) 1 x 1 2

Example 3 Find the length of the given curve. 2 f(x) 1 x

Example 3 – continued 2 f(x) 1 x

Lecture 4 – Surface Area Find the surface area by revolving the curve around the y-axis.

Find the surface area by revolving the curve around the x-axis.

Example 1 A: revolved about the y-axis. B: revolved about the x-axis. Put the surface areas in decreasing order for the given curve. A: revolved about the y-axis. B: revolved about the x-axis. C: revolved about the line x=2.

Example 1-continued 4 4 4 3 3 3 2 2 2 1 1 1

Lecture 5 – Physics Applications Mass 1D object: 3D object: If density varies along the length of the 1-D object (wires, rods), then use integrals to compute mass. a b

Example 1 Find the mass of the given wire.

Work require to move an object : If force varies along the way (horizontal, vertical), then use integrals to compute work. a b

Example 2 Find the work done by lifting a 28m chain (with mass density 2 kg/m) up to the top of a building.

Example 3 Suppose a force of 15 N is required to stretch and hold a spring a distance of .25 m past its natural length. Use Hooke’s Law to answer the following. What is the spring constant? How much work is done to compress the spring to .2 m from its natural length? If the spring is already stretched .2 m from equilibrium, then how much additional work is done to stretch it another .2 m?

Example 3 – continued a) b) c)

Example 4 Pump oil from the tank to the top of the tank. Oil has a density of 800 kg/m3 and the oil has a vertical depth of 10 m. Figure the work done to pump one slice: diam = 25 m 20 10 y

Lecture 6 – More Physical Applications Example 4 – continued Cross-sectional view: diam = 25 20 10

Hydrostatic Pressure is defined as the force per unit area Force and Pressure Hydrostatic Pressure is defined as the force per unit area of liquid exerted on an object. For any submerged object, the pressure depends only on depth.

Then, the force on an object submerged vertically (or the side of a dam or other construct) can be determined as: For a vertical object, find the force on one representative horizontal slice and let n go to infinity to make the Riemann sum become an integral.

Example 5 For the dam below, if water reaches to the top, then what force is exerted on the side of the dam? Water density is 1000 kg/m3. top = 400 m 220 base = 200 m

Example 5 – continued top = 400 m 220 base = 200 m

Example 6 A barrel of oil (half-full) is lying on its side. If each end is 8 feet in diameter, find the total force exerted by the oil against one of the sides. Oil has a weight density of 50 lb per square foot. 4

Example 6 – continued 4

Lecture 7 – Hyperbolic Functions Functions involving a combination of two exponentials. Applications – heat, electricity, catenary

Derivatives of hyperbolics

Example 1 What is the length of y = cosh x from x = -a to x = a?

Example 2 A power line hangs between two poles 14 m apart in the shape of a catenary y = 20 cosh (x/20) -15, where x and y are measured in meters. a) Find the slope where it meets the right pole. b) Find the angle between power line and pole.