Finding Volumes Chapter 6.2 February 22, 2007

In General: Vertical Cut: Horizontal Cut:

In-class Integrate: Set up an integral to find the area of the region bounded by:

Find the area of the region bounded by Bounds? In terms of y: [-2,1] Points: (0,-2), (3,1) Right Function? Left Function? Area?

Volume & Definite Integrals We used definite integrals to find areas by slicing the region and adding up the areas of the slices. We will use definite integrals to compute volume in a similar way, by slicing the solid and adding up the volumes of the slices. For Example………………

Blobs in Space Volume of a blob:

Blobs in Space Volume of a blob: Cross sectional area at height h: A(h)

Blobs in Space Volume of a blob: Cross sectional area at height h: A(h) Volume =

Example Solid with cross sectional area A(h) = 2h at height h. Stretches from h = 2 to h = 4. Find the volume.

Example Solid with cross sectional area A(h) = 2h at height h. Stretches from h = 2 to h = 4. Find the volume.

Example Solid with cross sectional area A(h) = 2h at height h. Stretches from h = 2 to h = 4. Find the volume.

Example Solid with cross sectional area A(h) = 2h at height h. Stretches from h = 2 to h = 4. Find the volume.

Volumes: We will be given a “boundary” for the base of the shape which will be used to find a length. We will use that length to find the area of a figure generated from the slice . The dy or dx will be used to represent the thickness. The volumes from the slices will be added together to get the total volume of the figure.

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square. [-1,1] Bounds? Top Function? Bottom Function? Length?

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square. We use this length to find the area of the square. Length? Area? Volume?

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square. What does this shape look like? Volume?

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a circle with diameter in the plane. Length? Area? Radius: Volume?

Using the half circle [0,1] as the base slices perpendicular to the x-axis are isosceles right triangles. Bounds? [0,1] Length? Area? Volume? Visual?

The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve. [0,π] Bounds? Top Function? Bottom Function? Length? Area of an equilateral triangle?

Area of an Equilateral Triangle? Area = (1/2)b*h

The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve. [0,π] Bounds? Top Function? Bottom Function? Length? Area of an equilateral triangle? Volume?

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane. We used this length to find the area of the square whose side was in the plane…. Length? Area with the length representing the diagonal?

Area of Square whose diagonal is in the plane?

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane. Length of Diagonal? Length of Side? Area? Volume?

Solids of Revolution We start with a known planar shape and rotate that shape about a line resulting in a three dimensional shape known as a solid of revolution. When this solid of revolution takes on a non-regular shape, we can use integration to compute the volume. For example……

The Bell! Area: Volume:

Find the volume of the solid generated by revolving the region defined by , x = 3 and the x-axis about the x-axis. Bounds? [0,3] Length? (radius) Area? Volume?

Find the volume of the solid generated by revolving the region defined by , x = 3 and the x-axis about the x-axis.

Solids of Revolution Rotate a region about an axis.

Solids of Revolution Rotate a region about an axis. Such as: Region between y = x2 and the y-axis, for 0 ≤ x ≤ 2, about the y-axis.

Solids of Revolution Rotate a region about an axis. Such as: Region between y = x2 and the y-axis, for 0 ≤ x ≤ 2, about the y-axis.

Solids of Revolution For solids of revolution, cross sections are circles, so we can use the formula Volume =

Solids of Revolution For solids of revolution, cross sections are circles, so we can use the formula Usually, the only difficult part is determining r(h). A good sketch is a big help. Volume =

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. We need to figure out what the resulting solid looks like.

Aside: Sketching Revolutions Sketch the curve; determine the region.

Aside: Sketching Revolutions Sketch the curve; determine the region. Sketch the reflection over the axis.

Aside: Sketching Revolutions Sketch the curve; determine the region. Sketch the reflection over the axis. Sketch in a few “revolution” lines.

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. Now find r(y), the radius at height y.

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. Now find r(y), the radius at height y. (Each slice is a circle.) y

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. y

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. x = sin(y)

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. So r(y) = sin(y).

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. So r(y) = sin(y). Therefore, volume is

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. So r(y) = sin(y). Therefore, volume is

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. So r(y) = sin(y). Therefore, volume is What are the limits?

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. So r(y) = sin(y). Therefore, volume is What are the limits? The variable is y, so the limits are in terms of y…

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. Upper limit: y = π Lower limit: y = 0

Example Rotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis. Volume is

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. (Since we rotate about x, the slices are perpendicular to x. So x is our variable.)

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. First, get the region:

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. First, get the region:

When we rotate, this will become a radius. Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. First, get the region: When we rotate, this will become a radius.

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Next, revolve the region:

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Next, revolve the region:

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Radius at point x is 3e–x.

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Radius at point x is 3e–x. Limits are from x =  to x = 

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Radius at point x is 3e–x. Limits are from x = 0 to x = 1

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Radius at point x is 3e–x. Limits are from x = 0 to x = 1 Volume:

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Radius at point x is 3e–x. Limits are from x = 0 to x = 1 Volume:

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Radius at point x is 3e–x. Limits are from x = 0 to x = 1 Volume:

Example Revolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis. Radius at point x is 3e–x. Limits are from x = 0 to x = 1 Volume:

Remember for this method: Slices are perpendicular to the axis of rotation. Radius is then a function of position on that axis. Therefore rotating about x axis gives an integral in x; rotating about y gives an integral in y.

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis.

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. First, sketch:

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, rotate about the x axis:

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, rotate about the x axis:

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = 

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = 

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = x2  Limits:

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = x2  Limits: x = 0 to x = 4

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = x2  Limits: x = 0 to x = 4 Volume:

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = x2  Limits: x = 0 to x = 4 Volume:

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = x2  Limits: x = 0 to x = 4 Volume:

Example Rotate y = x2, from x = 0 to x = 4, about the x-axis. Next, find r(x): r(x) = x2  Limits: x = 0 to x = 4 Volume:

Find the volume of the solid generated by revolving the region defined by , on the interval [1,2] about the x-axis. Bounds? [1,2] Length? (radius) Area? Volume?

Find the volume of the solid generated by revolving the region defined by , on the interval [1,2] about the x-axis.

Find the volume of the solid generated by revolving the region defined by , y = 8, and x = 0 about the y-axis. Bounds? [0,8] Length? Area? Volume?

Find the volume of the solid generated by revolving the region defined by , and y = 1, about the line y = 1 Bounds? [-1,1] Length? Area? Volume?

(Temporarily end powerpoint Look at Washer2.gif Washer5.gif What if there is a “gap” between the axis of rotation and the function? (Temporarily end powerpoint Look at Washer2.gif Washer5.gif Washer8.gif Washer10.gif)

*Find the volume of the solid generated by revolving the region defined by , and y = 1, about the x-axis. Bounds? [-1,1] Outside Radius? Inside Radius? Area? Volume?

Find the volume of the solid generated by revolving the region defined by , and y = 1, about the line y=-1. Bounds? [-1,1] Outside Radius? Inside Radius? Area? Volume?

Try: Set up an integral integrating with respect to y to find the volume of the solid of revolution obtained when the region bounded by the graphs of y = x2 and y = 0 and x = 2 is rotated around a) the y-axis b) the line x = 4

Summary Described how a plane region rotated about an axis describes a solid.

Summary Described how a plane region rotated about an axis describes a solid. Found volumes of solids of revolution using the area of a circle

Summary Described how a plane region rotated about an axis describes a solid. Found volumes of solids of revolution using the area of a circle Determined the variable as given by the axis of rotation.

Solids of Revolution If there is a gap between the function and the axis of rotation, we have a washer and use: If there is NO gap, we have a disk and use: Volume = Volume =