and y 3 and find the volume of the solid produced when that region is revolved a

Question

and y 3 and find the volume of the solid produced when that region is revolved about the r-axis (2) Sketch the region enclosed by y and y z, and find the volume of the solid produced when that region is revolved about the y axis. (3) sketch the region in the first quadrant enclosed by y r and y r and find the volume of the solid produced when that region is revolved about the line T 2. (4) Use an integral to find the volume of the solid with base the region enclosed by the graphs of y r4 and y 1 and whose cross sections perpendicular to the y-axis are squares. Draw a picture! You will need to determine the formula for the area of cross-sections to apply the integral formula for volume.

Explanation / Answer

1) In this problem we will apply shell method
y = 2x + 3 ----> (y - 3)/2 = x
y = x^2 ----> x = +/- (y)

height(1) = (y) - -(y) ----> 2(y) from 0 to y(-1) or 1
height(2) =(y) - (y - 3)/2 from y(-1) or 1 to y(3) or 9
radius =y

1. . . . . . . . . . .. . . 9
2 * y * 2(y) dy + 2 * y * ( (y) - (y - 3)/2 ) dy = (1088/15)   Answer

0                            1

2) For this problem we will try by using washer method

Using the Washer Method:
y = (5/2) - x
x = (5/2) - y

y = 1/x ---> x = 1/y

limits:
when x = 1/2 ---> y = 2
when x = 2 ---> y = 1/2

A (y) = ( outer radius )^2 - ( inner radius )^2
A (y) = ( (5/2) - y - 0 )^2 - ( 1/y - 0 )^2
A (y) = ( (25/4) - 5y + y^2 ) - ( 1/y^2 )
A (y) = ( (25/4) - 5y + y^2 - 1/y^2 )

2
( (25/4) - 5y + y^2 - 1/y^2 ) dy = 9/8
1/2

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