Finding Volumes. In General: Vertical Cut:Horizontal Cut:

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Presentation transcript:

Finding Volumes

In General: Vertical Cut:Horizontal Cut:

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! Volume: Area:

Solids of Revolution Rotate a region about an axis. Such as: Region between y = x 2 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 = x 2 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 Usually, the only difficult part is determining r(h). A good sketch is a big help. 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? Length? (radius) Area? Volume? [0,3]

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 1.Sketch the curve; determine the region. 2.Sketch the reflection over the axis. 3.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. 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). 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. 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. 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 fromx = 0to 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 = x 2, from x = 0 to x = 4, about the x-axis. Next, find r(x):r(x) = x 2 Limits: x = 0 to x = 4

Example Rotate y = x 2, from x = 0 to x = 4, about the x-axis. Next, find r(x):r(x) = x 2 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? Length? (radius) Area? Volume? [1,2]

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

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

Find the volume of the solid generated by revolving the region defined by, and y = 1, about the x-axis using planar slices perpendicular to the AOR.

Area of a Washer Note: Both R and r are measured from the axis of rotation.

Solids of Revolution: We determined that a cut perpendicular to the axis of rotation will either form a disk (region touches axis of rotation (AOR)) or a washer (there is a gap between the region and the AOR) Revolved around the line y = 1, the region forms a disk However when revolved around the x-axis, there is a “gap” between the region and the x-axis. (when we draw the radius, the radius intersects the region twice.)

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

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

Let R be the region in the x-y plane bounded by Set up the integral for the volume obtained by rotating R about the x-axis using planar slices perpendicular to the axis of rotation.

Notice the gap: Outside Radius ( R ): Inside Radius ( r ): Area: Volume:

Let R be the region in the x-y plane bounded by Set up the integral for the volume of the solid obtained by rotating R about the x-axis, using planar slices perpendicular to the axis of rotation.

Notice the gap: Outside Radius ( R ): Inside Radius ( r ): 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? Length? (radius) Area? Volume? [0,3]