Thread: Cylindrical Shell Method + other Volume question

Having trouble with these two problems:

1) Region bounded by y= sqrt[x], y = 2, x =0

Find the area using the shell method by revolving the enclosed region around the following axis:
a) x-axis
b) y-axis
c) x=4
d) y=2


2) Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve y=x sin x, 0≤x≤pi about,
a) y -axis
b) the line x= pi

Thanks!!

Originally Posted by Jeffman50

Having trouble with these two problems:

1) Region bounded by y= sqrt[x], y = 2, x =0

Find the area using the shell method by revolving the enclosed region around the following axis:
a) x-axis

A line parallel to the x-axis at distance y from the x- axis would go from to and, rotated around the x-axis would form a cylinder of height and radius y. A circle around the x-axis with radius y has circumference and so the cylinder has surface area . Taking dy as the thickness, the volume of the thin shell would be and the entire volume, .

b) y-axis

It would be easier to use the "disk" method but since you specifically ask about the shell method: Rotating around the y-axis, a line parallel to the y-axis would go from to y= 2, so have length and have radius x. Its surface area is . The volume of such a thin shell is [tex]2\pi x(2- \sqrt{x})dx and the entired area is .

c) x=4

Instead of the radius being "x" as in (b), the radius would be "4- x".

d) y=2

Instead of the radius being "y" as in (a), the radius would be "2- y".

2) Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve y=x sin x, 0≤x≤pi about,
a) y -axis

Using shells, each shell would have height xsin(x)- 0= xsin(x) and the radius would b x. The volume is .

b) the line x= pi

Same as a except the radius is .

Thanks!![/QUOTE]