7.1 Area Between 2 Curves Objective: To calculate the area between 2 curves. Type 1: The top to bottom curve does not change. a b f(x) g(x) *Vertical rectangles top – bottom gives the height This will sum the rectangles from the left to the right with width Dx. Ex: Find the area bounded by y = x and y = 4x – x2.

Type 2: The top to bottom curve change within the given interval. f(x) g(x) a b c If the top to bottom curve change within the interval, you must start a new integral. You will also need to calculate the point of intersection of the 2 curves!!! Ex: Find the area bounded by the curves y = x2 and y = 2 - x2 from [-2, 1].

Type 3: If you need to use horizontal rectangles…. g(y) c d * Now (right – left) gives height and all equations must be in terms of y. You will also integrate from the bottom to the top. You will also need the y-value of the point of intersection. Ex: Find the area with respect to the y-axis bounded by the curves y = 2x and y = x3 from y = -2 to y = 1. Could you calculate this area to the x-axis?

Ex: Find the area bounded by  a) y = x2 + 1 and y = 2 b) x = y2 and y = x - 2 Can you calculate the area between 2 curves? Assign 7.1: Set up each integral, then use your calculator to check your answer for each: 1-6, 17, 18, 19, 21, 25, 27, 31, 33 odd, 37-41odd, 61, 69, 73, 105

7.2a Volumes of Revolution Objective: To calculate the volume of a solid when a region is revolved about the x or y-axis.  (Disk Method) Disks to the x-axis: This means you are revolving your region about the x-axis. If f(x) is revolved about the x-axis the solid created has solid circular cross sections with radius = f(x). Rectangle   to the x-axis  this is the radius of the circle which is the height of the curve, f(x). Integrate from a to b (left to right) along the x-axis f(x) a b

Ex: Find the volume created when y = x2 is revolved about the x-axis from [0, 2].

Disks to the y-axis: This means you are revolving your region about the y-axis. If f(x) is revolved about the y-axis the solid created has solid circular cross sections with radius = f(y). Rectangle   to the y-axis  this is the radius of the circle which is the height of the curve, f(y). Integrate from c to d (bottom to top) along the y-axis

Ex: Find the volume created when y = x2 is revolved about the y-axis from y = 0 to y = 4. Note: This is the same equation as the previous example, but when spun about the y- axis creates a different volume.

Washer Method: This is when you are revolving a region that is not bounded to the x or y-axis. Your cross section looks like a washer. What is the area of the blue shaded region? Area = Area of large circle – Area of small circle A = pR2 – pr2 (Outer circle – Inner Circle) Find the volume when the region bounded by y = x2 + 1, y = x [0, 2] is revolved about the x-axis. x

Use what you have learned to derive the formula for the volume of a sphere with radius r. Hint: Let the center be (0, 0) and rotate about the x-axis. What shape do you need? What is the equation for this shape? Make sure you know the difference between the disk and washer method. You must also know the difference in the set-up depending upon the axis you revolve about (what variables should you use?)

Other Axes of Revolution Objective: To determine the volume when the region is spun about any horizontal or vertical axis. Spin about y = c: Remember, you must look at the outer2 – inner2 with respect to the axis of revolution. 1. Above the region 2. Between 3. Opposite side of axis y = c a b

NOTE: The axis of revolution cannot pass through the region you plan to revolve. If you spin about a horizontal axis, all variables must be in terms of … Find the volume when you spin the region enclosed by y = x2 , y = 0 and x = 2 about each horizontal line. (Set up the integral,then use your calculator.) a) y = 8 b) y = 4 c) y = -2

Spin about x = c: Remember, you must look at the outer2 – inner2 Spin about x = c: Remember, you must look at the outer2 – inner2 with respect to the axis of revolution. When you spin about a vertical axis, all variables must be in terms of … Find the volume when you spin the region enclosed by y = x2 , y = 0 and x = 2 about each vertical line. (Set up the integral,then use your calculator.) a) x = 4 b) x = 2 c) x = -2

Find the volume when the region bounded by y = 6 – 2x – x2 and y = x + 6 is revolved about a) x-axis b) y = 3 c) y = 7 Can you set up volumes to other axes of revolution? Do you know what variables to use for horizontal/vertical axes of revolution? Set up / Calculator / check answer  Don’t forget p! 7-2a: 1-22 all (Set up each integral, then use your calculator to check each answer.)

7.2b Volumes by Slicing  Questions from 6.2a or 6.2b? Objective: To find the volume of a solid with a known cross section by slicing. Questions from 6.2a or 6.2b? Find the volume of the solid whose base is bounded by y = 2x and y = x2  cross sections perpendicular to the x-axis are a) squares: b)semi-circles: c) equilateral triangles: d) rectangles with height = 2: e) isosceles right triangles with leg = base: f) isosceles right triangles with hypotenuse = base: base Why is it important to know that the cross sections are  to the x-axis? Note: These do not spin! You will not have p unless it is in the area formula.

squares: b) semi-circles: base = base = A of cross section = A of cross section = V = V = What is the base of the solid? (Sketch the base of the solid and determine the length of the base of each slice.) What is the area of the cross section in terms of the base? V =

equilateral triangles: d) rectangles with height = 2: base = base = A of cross section = A of cross section = V = V =

e) leg = base: f) hypotenuse = base: Isosceles right triangles  e) leg = base: f) hypotenuse = base: base = base = A of cross section = A of cross section = V = V =

squares: rectangles with height = 2: There is a short cut here. semi-circles: equilateral triangles: rectangles with height = 2: e) isosceles right triangles with leg = base: f) isosceles right triangles with hypotenuse = base:

Cross Sections on Circular Bases Find the volume of the solid whose base is the region bounded by x2 + y2 = 4 if cross sections  to the x-axis are  This is the base of your shape. Find the base in terms of x if  the x-axis and y if  to the y-axis. A = V = a) Squares A = V = b) isosceles right -leg base

You try these! Find the volume of the solid whose base is the region bounded by x2 + y2 = 64 if cross sections  to the x-axis are  a) Equilateral triangles b) rectangles with h = 3 Can you determine volumes by slicing? Do you know when you need p? 7-2b: 27, 57-59**, 62a, 63, 71, 72, 74, 75

7.4 Arc Length and Surfaces of Revolution Objective: To calculate the arc length of a smooth curve and determine the area of a surface of revolution. Let f(x) represent a smooth curve on the interval [a, b]. The arc length of f(x) between a and b is : Proof: page 478 or Find the arc length of the function between the points (1, 1) and (16, 4).

Which function is longer on the interval [0, 2]: f(x) = x2 or g(x) = x3 Surface of Revolution : If a curve is revolved about a line, the result is a surface of revolution. Where r(x) is the distance between the curve and the axis of revolution. or  this formula sums 2r (the circumference) for each circle - 

Find the surface area generated when f(x) is revolved about the x-axis. r(x) = Ans: 8p You will need these again when we do parametric and polar equations.  Assign: 7.4: 3-11odd, 17-21odd, 32, 37-39, 55, 57, 61a, b, c

Know how to calculate area to the x and y axes Ch 7 Review ~ WS 1 & WS 2 Know how to calculate area to the x and y axes Know how to set up volumes for the disc and washer method to any horizontal or vertical axis. Spin about y = c: Remember, you must look at the outer2 – inner2 with respect to the axis of revolution.

Know how to determine volumes by slicing. If the cross sections are: a) squares: b) semi-circles: c) equilateral triangles: d) rectangles with height = 2: e) isosceles right triangles with leg = base: f) isosceles right triangles with hypotenuse = base:

Arc Length for a smooth curve: Surface of revolution: Test Tomorrow 