Here is quiz 4.
Problem 1 is Exercise 7.1.36 from the book. Here is a graph showing the curve for
We rotate about the line the region bounded by the -axis and this curve.
To find its volume, a first natural attempt would be to use the washer method. We would then attempt to compute the volume as
where is the maximum of for and, for any given value of with and are the values of with such that
Unfortunately, this does not seem to be the best approach, as there does not seem to be a reasonable way of solving for the equation
(In fact, this equation cannot be solved in terms of elementary functions.
Similarly, there is no way of finding exactly what the value of is in terms of elementary functions.)
Since this approach seems to lead us nowhere, we now try to compute the volume using the shell method. Now the volume is expressed as
This expression looks approachable with the techniques we have studied. First, let’s rewrite the integral as
We compute both expressions using integration by parts:
To find we use and so and we can take Hence
We recognize from the graph of that the second expression is zero, and we have:
Similarly, for we have and so and and
The last expression is once more computed using parts, now with and so and This gives
Finally, the required volume is
Problem 2 asked to evaluate
A first attempt may go by using integration by parts, with and Unfortunately, this approach would not lead to simpler expressions, as both integrals and derivatives of and carry radicals.
If the expression inside the square root were of the form or we could use a trigonometric substitution. However, is not of this form. On the other hand, in Chapter 9 we saw that it is sometimes useful to complete squares, so we may want to try that here. We have:
This suggest trying the trigonometric substitution for We have and Also, when we have or and when we have or
In terms of the integral becomes
To evaluate expressions of this form, we use the identity and obtain
The second expression we recognize as For the first, we use either the reduction formula found in lecture, or integration by parts:
from which we get
so the required integral equals
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