Question 4

Let Q be the shaded region in the first quadrant enclosed by the graph y= 2cos(x), and y=e^x2 and the y-axis as shown in the figure.



a) Find the area of region Q.

b) Find the volume of the solid generated when Q is revolved about the x-axis.

c) The region Q is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of this solid.


Solution

a) Answer, [0 ∫ a] (2cosx - e^x2 ) dx = 0.4566

For these questions you find which graph is furthest on top visually, and subtract it with the bottom graph.

• You would also need to find the points of intersections to be able to find the area in between.
x-intersects at x=0 and x=0.67031574 when 2cosx=e^(x2). Let a=0.67031574.
In this case y=2cosx is on top and y=e^x2 is on the bottom.

• By taking the integral of (2cosx)-(e^x2) we get the area of Q.

b) Answer, Volume = π [ 0 ∫ a ] ( (2cosx)² - (e^x2)² dx = 1.3751

Visually picturing the area being revolved around the x-axis.

• We see that it makes washers (Circular figure with a hole in the center).



• The area of the washer at one a particular point can be expressed as πR² - πr² where R is the bigger radius and r is the smaller one.
Factoring out pi, we get π(R² - r²). In this case the bigger radius is (2cosx - 1), where as the smaller radius is (e^x2 - 1)².

• By taking the integral of π(R² - r²), will give you the total area over the designated time interval.
Keep in mind that the interval starts from 0 because we are only worried about the shaded region highlighted in the first quadrant. If it were to be the whole area in between the graphs then that would be from -a to a.

c) Answer, Volume = [ 0 ∫ a ] [2cosx - e^x2]² dx = 0.3687

From the picture we can clearly see what the question is asking.



• By taking the integral of the area of the base of the solid squared, will give you the volume of the solid.
The solid generated is a cube.