壳层法计算旋转体体积的方法求助

Hey there! Let's break this down step by step—you've got a classic case of mixing up the setup for one of the methods, and it's totally easy to do when first learning these rotation volume techniques.

First off, your shell method setup is actually correct! Let's re-calculate it to confirm:
2π∫₀² x(4-x²) dx
Expand the integrand first: 4x - x³
Integrate term by term: 2π[2x² - (x⁴)/4] evaluated from 0 to 2
Plug in the bounds: 2π[(2*(2)² - (2⁴)/4) - 0] = 2π[(8 - 4)] = 8π
That's the right result here.

Now, the issue is that your disk method result of 6π is incorrect—you must have messed up the setup for that method. Let's walk through the correct disk method approach for rotating around the y-axis:
When using the disk method for rotation around the y-axis, we need to integrate with respect to y (not x). First, rewrite the function y=4-x² in terms of x: x=√(4-y) (since we're dealing with x ≥ 0 in the interval [0,2]).
The disk radius at any y is x=√(4-y), so the area of each disk is π*(radius)² = π*(4-y).
The bounds for y go from 0 (when x=2) up to 4 (when x=0).
So the integral is:
π∫₀⁴ (4-y) dy
Integrate that: π[4y - (y²)/2] evaluated from 0 to 4
Plug in the bounds: π[(4*4 - (4²)/2) - 0] = π[(16 - 8)] = 8π
That matches the shell method result, which makes sense—both methods should give the same volume when set up correctly.

My guess is that you accidentally set up the disk method for rotation around the x-axis instead of the y-axis, or integrated with respect to x incorrectly. For example, if you tried an integral that didn't account for the axis of rotation properly, that would lead to a wrong result like 6π.

So to wrap up: your shell method was perfect all along! The mistake was in your disk method setup. Double-checking which axis you're rotating around and making sure your integral variable matches the method (shell uses x for y-axis rotation, disk uses y for y-axis rotation) will help avoid this kind of mix-up in the future.

备注：内容来源于stack exchange，提问作者Bac Dang