TYPES OF SLICES (OR "CROSS-SECTIONS")
When finding the volume by slicing, we end up with several different shapes of slices. Below are the common names for these slices.
Small cylinders stacked together, such as in the Lime Lab. The volume of each disk can be found with the cylinder formula V = πr2 h , where ris the radius and h is usually ∆x or ∆y .
These are disks with holes in the middle, such as in problem 8-25. The volume of each washer can be found using V = π(R2 − r2 )h where R is the outer radius, r is the inner radius, and h is usually Δx or Δy.
This general name applies to all other slices that have non-circular cross-sections, such as the Goodslice bread example. The base of each prism could be triangular, square, rectangular, trapezoidal, etc. The volume of each prism can be generally found by using V = (base area)(height) where height is usually Δx or Δy.
When finding volumes of solids, be sure to include the following:
1. A rough sketch of the solid.
2. A typical slice with dimensions labeled.
3. The method used (disk or washers).
4. The integral that will compute the volume.
5. The volume found by evaluating the integral.
Arc length If f(x) is differentiable on [a, b], then the distance along the curve from x = a to x = b is given by the integral:
The basic process for solving any related rates problem is:
Use a diagram and geometry to find a relationship between the changing quantities. For example, is there an appropriate area or volume formula that you could apply? Is there a right triangle for which the Pythagorean Theorem could be applied? Are there similar triangles?
Use implicit differentiation to find a rate relationship between the terms. Do not forget to differentiate with respect to time (if time is important)!
Substitute quantities and rates to solve for the desired rate. Be sure to make any decreasing rates negative.
Can you tell just by looking at the integral at right that it is a "reverse Chain Rule" problem? How?
Let's assume the inside function is . To clean up the messy integral, let's call the inside function u, and substitute that into the integral. Oh-oh! Now we have u' s, x's and dx' s in the same integral. We need to convert everything to u' s. To replace dx, we can differentiate our u equation and solve for dx. Now we can substitute this expression in for dx in the integral: If we chose our u correctly, the remaining x' s should disappear through simplification. At last! Now we can write the integral completely in terms of u: Now we can integrate:
And finally, since , we can substitute this back in for u and simplify: olving Differential Equations
The equation is called a differential equation because it involves the derivative (or rate of change). We have…