5. Set up, but do not evaluate, an integral which gives the volume of the follow

Question

5. Set up, but do not evaluate, an integral which gives the volume of the following figure: S is a 3-D figure with base a circle of radius 1. Cross sections of the region taken perpendicular to the base are squares. 6. Let R be the region bounded by (x -1)^2 + y^2=1 where y >= 0, and x>= 1. Consider the volume R formed by rotating the region about the line x = 2. Set up, but do not solve, an integral of the form integral f(x) dx which gives the volume of this region. (That is, x must be your integration variable!) 7. Repeat the previous exercise giving an integral of the type R integral f(y)dy. 8. Consider the curve y = e^-x with x >=0. Take the surface are of the figure created by rotating this curve about the x-axis. Is this area finite or infinite? Give a complete reason for your conclusion. 9. Derive the arc length formula for a curve given by a continuous function y = f(x).

Explanation / Answer

Finding the volume of a solid revolution is a method of calculating the volume of a 3D object formed by a rotated area of a 2D space. Finding the volume is much like finding the area, but with an added component of rotating the area around a line of symmetry - usually the x or y axis. (1) Recall finding the area under a curve. Find the area of the definite integral Integrate across [0,3]: Now, let's rotate this area 360 degrees around the x axis. We will have a 3D solid that looks like this: To find this volume, we could take vertical slices of the solid (each dx wide and f(x) tall) and add them up. This is quite tedious, but thankfully we have calculus! Since the integrated area is being rotated around the axis under the curve, we can use disk integration to find the volume. Since the area is rotated full circle, we can use the formula for area of a cylinder to find our volume. Volume of a cylinder We can merge the formula for volume of a cylinder and our definite integral to find the volume of our solid. The radius for our cylinder would be the function f(x) and the height of our cylinder would be the distance of each disk: dx The volume of each slice would be Adding the volumes of the disks with infinitely small dx would give us the formula Using our function, we would get this integrand for the volume Evaluating the integral, we get We obtained 4? units3 as our volume. Since our function is linear and the radius is changing at a constant rate, it is easy to check this by plugging in values to the formula for volume of a cone. The answers are the same. Since our function was linear and shaped like a cone when rotated around the x axis, it was okay to use the volume formula for a cone. Many of the volumes we will be working with are not shaped like cone, so we cannot simply substitute values in the formula. While algebra can take care of the nice straight lines, calculus takes care of the not-so-nice curves.

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