What Is Wrong with My Washer Method Calculation?

[talk2glenn]

Homework Statement

The region bounded by x = 1 - y⁴, is rotated about the line y = -4.
The volume of the resulting solid is:

Homework Equations

Area of a circle: \pir²
Washer method solution: \int_a^{b} pi*[f(x)^2-g(x)^2]

The Attempt at a Solution

I divided the covered region into two circles: outer with R = 4 + y and inner with r = 4. Then solved using washer method, on the integral from 0 to 1 with respect to x.

\pi\int_0^{1} [4+(1-x)^{1/4}]^2-4^2dx = 106\pi/15

Told this is wrong by the computer. Very frustrating, as I can see no other way to set up this integral, having drawn the diagrams. Any idea what I'm doing wrong? Thanks in advance :)

 

[phinds]

I get the R = y + 4, but what's the R = 4?

What I get for the solid is an upside-down cake turned on its side and with the bottom extended forever so I get an infinite volume that is curve-cone shaped with an indentation in the "top" (the part starting at the level of x=1 and closing up at x=-255). Haven't done math in so long that I have no idea what the washer method is so can't help you with that.

 

[talk2glenn]

You seem to have the shape correct, but the function is undefined at y = 0 (a global min), so practically speaking the x-axis is an unspoken lower bound. This is why I subtract an inner circle of radius 4.

Otherwise, absent this lower bound, you do indeed have an infinite volume beneath the curve. There is also a given bound at x=0 that I didn't mention in the setup.

Does that clear it up?

 

[phinds]

I don't understand your statement that there is a lower bound on the function but no matter because even with that lower bound the volume is still infinite and I don't see why you say it is not? You have an infinite volumn with a cylinder cut out of the middle, so it is still an infinite volume. Am I missing something?

 

[SammyS]

Why do you integrate from 0 to 1 ? The problem makes no statement about this.

Is there more to the problem than you have stated?