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VOLUMES OF SOLIDS OF REVOLUTION

VOLUMES OF SOLIDS OF REVOLUTION. Many solid objects, especially those made on a lathe , have a circular cross-section and curved sides. In this section, we see how to find the volume of such objects using integration. Objects made on a lathe.

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VOLUMES OF SOLIDS OF REVOLUTION

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  1. VOLUMES OF SOLIDS OF REVOLUTION Many solid objects, especially those made on a lathe, have a circular cross-section and curved sides. In this section, we see how to find the volume of such objects using integration. Objects made on a lathe ...

  2. Consider the area bounded by the straight line y = 3x the x-axis, and x = 1: When the shaded area is rotated 360° about the x-axis, we observe that a volume is generated. The resulting solid is a cone:

  3. Disk Method for Finding Volumes To find this volume, we could take slices (the yellow disk shown above), each dx wide and radius y: The volume of a cylinder is given by: V = πr2h Because radius = r = y and each disk is dx high, we notice that the volume of each slice is: V = πy2 dx Adding the volumes of the disks (with infinitely small dx), we obtain the formula: which means 

  4. where: y = f(x)is the equation of the curve whose area is being rotated a and b are the limits of the area being rotated dx shows that the area is being rotated about the x-axis Applying the formula to the earlier example, we have:

  5. Example: (Practical Application) A watermelon has an ellipsoidal shape with major axis 28 cm and minor axis 25 cm. Find its volume. Historical Approach: Before calculus, one way of approximating the volume would be to slice the watermelon (say in 2 cm thick slices) and add up the volumes of each slice using V = πr2h. Interestingly, Archimedes (the "Eureka! I've got it" dude) used this approach to find volumes of spheres around 200 BC. The technique was almost forgotten until the early 1700s when calculus was developed by Newton and Leibniz. We see how to do the problem using both approaches.

  6. Volume using historical method: Because the melon is symmetrical, we can work out the volume of one half of the melon, and then double our answer. The radii for the slices for one half of a particular watermelon are found from measurement to be: 0, 6.4, 8.7, 10.3, 11.3, 12.0, 12.4, 12.5. The approximate volume for one half of the melon using slices 2 cm thick would be: So the volume for the whole watermelon is about 5054.4 × 2 = 10109 cm3 = 10.1 L.

  7. "Exact" Volume (using Integration): We are told the melon is an ellipsoid. We need to find the equation of the cross-sectional ellipse with major axis 28 cm and minor axis 25 cm. We use the formula : where a is half the length of the major axis and b is half the length of the minor axis. For the volume formula, we will need the expression for y2 and it is easier to solve for this now (before substituting our a and b). Since a = 14 and b = 12.5, we have: NOTE: The a and b that we are using for the ellipse formula are not the same a and b we use in the integration step. They are completely different parts of the problem.

  8. Using this, we can now find the volume using integration. (Once again we find the volume for half and then double it at the end). So the watermelon's total volume is 2 × 4580.65 = 9161 cm3 or 9.161 L. This is about the same as what we got by slicing the watermelon and adding the volume of the slices.

  9. Summary of the Disk Method: where: y = f(x)is the equation of the curve whose area is being rotated a and b are the limits of the area being rotated dx shows that the area is being rotated about the x-axis

  10. The Washer Method

  11. Washer Method: If R is the radius of the outer disk and r is the radius of the inner disk, then the area of the washer is , and its volume would be its area times its thickness.

  12. Volume by washers perpendicular to X-axis

  13. Example 1 (Draw diagrams) The line joining the points (0,2) and (4,6) is rotated about the x-axis. Find the volume of the solid of revolution. Example 2 A hole of radius “a” is bored through the centre of a sphere of radius “2a”. Find the volume of the remaining solid.

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