volume of the solid

ProBasket

The region bounded by [tex]x = 1- y^4, x=0 [/tex] is rotate about the line [tex] x = 3 [/tex] The volume of the resulting solid is ......


here's what i done:
[tex]x = 1- y^4 [/tex] in terms of y => [tex] y = (1-x)^{1/4}[/tex]


my integral:
[tex]\int_0^{1} 2*pi*(3-x)(1-x)^{1/4}[/tex]

anyone know what i have done incorrectly?

ehild

This is a hollow body, you have to subtract the volume of the body with radius r = 3- x(y) in the picture from that of the cylinder of radius R=3. And integrate with respect to y as the rotational angle is parallel to the y axis.

[tex]V=\pi \int_{-1}^1 {(3^2-r^2)dy} [/tex]

ehild

BobG

Normally, either the shell method or the washer method will work. In this case, you're going to have some problems with [tex] (1-x)^{\frac{1}{4}}[/tex].

It's not a function since an even root has both a positive and negative root. For example:

[tex]\sqrt {1-.84}= \pm .4[/tex]

Without the bottom half of the graph, you can't find the volume. You could compensate just by finding the volume of the region bounded by x=(1-x)^(1/4), x=0, y=0, and then multiplying by two.

It would be easier to just to use the washer method, as ehild suggested.