Volume of Solid Rotated About x=3, Region Bounded by x=1-y^4, x=0

[ProBasket]

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


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


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

anyone know what i have done incorrectly?

 

[ehild]

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


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


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

anyone know what i have done incorrectly?

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.

$$V=\pi \int_{-1}^1 {(3^2-r^2)dy}$$

ehild

 

[BobG]

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


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


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

anyone know what i have done incorrectly?

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

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

$$\sqrt {1-.84}= \pm .4$$

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.