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# Question 3

last edited by 16 years, 5 months ago

# Question 3

## You MAY use a calculator to solve this problem.

Let R be the region enclosed by the graphs of y = 2ln(x) and y = x/2, and the lines x = 2 and x = 8.

(a) Find the area of R.

(b) Set up __but do not integrate__, an integral expression, in terms of a single variable, for the volume of the solid generated when R is revolved about the x-axis.

(c) Set up __but do not integrate__, an integral expression, in terms of a single variable, for the volume of the solid generated when R is revolved about the line x = -1.

Solution

A) We are finding the area bounded by the four graphs of y = 2ln(x) and y = x/2, and the lines x = 2 and x = 8.

The area bounded by the graphs is found on integral from 2 to 8.

The area R is the total area under the curve of the TOP function minus the total area under the curve of the BOTTOM function. In this case the top function is y=2ln(x) and the bottom function is y=x/2. ********************************************************************

--There's another way you can find the area of R on your calculator

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B) To find the volume of the solid, think of the shape that each function will make as it is rotated around the x-axis. If you look at a very small change in x, a side view of the solid will look like a circle. The lower function will result in a hole within each circle, forming a washer-type shape. So the volume can be found by adding up all the little areas of the washers. It is important to note that the area of a circle is (pi)r^2. The radii of the inner and outer circles are their respective functions (outer: 2ln(x) and inner: x/2). By integrating on the interval (2,8) the function described by the area of the larger circle minus the area of the smaller circle, you can find the volume of the solid R. C) 