# Thread: Cylindrical Shell Method + other Volume question

1. ## 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!!

2. 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 $\displaystyle (0, y)$ to $\displaystyle (y^2, y)$ and, rotated around the x-axis would form a cylinder of height $\displaystyle y^2$ and radius y. A circle around the x-axis with radius y has circumference $\displaystyle 2\pi y$ and so the cylinder has surface area $\displaystyle 2\pi y(y^2)= 2\pi y^3$. Taking dy as the thickness, the volume of the thin shell would be $\displaystyle 2\pi y^3 dy$ and the entire volume, $\displaystyle 2\pi\int_{y= 0}^2 y^3 dy$.

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 $\displaystyle y= \sqrt{x}$ to y= 2, so have length $\displaystyle 2- \sqrt{x}$ and have radius x. Its surface area is $\displaystyle 2\pi x(2- \sqrt{x}$. The volume of such a thin shell is [tex]2\pi x(2- \sqrt{x})dx and the entired area is $\displaystyle 2\pi\int_0^4 x(2- \sqrt{x})dx= 2\pi\int_0^4 (2x- x^{3/2})dx$.

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
Using shells, each shell would have height xsin(x)- 0= xsin(x) and the radius would b x. The volume is $\displaystyle 2\pi\int_0^\pi x(x sin(x)dx= 2\pi\int_0^\pi x^2 sin(x) dx$.

b) the line x= pi
Same as a except the radius is $\displaystyle \pi- x$.

Thanks!![/QUOTE]

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# rotate the region bounded by y=sqrt(x), y=3 about the y axis

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