Why can't I integrate with respects to x for this volume of integration problem using shell's method?

Question

Here is the graph of the problem, http://gyazo.com/5755dcfeea517756762943f704280d91

Here is the problem itself, http://gyazo.com/14bf36a0fa1968bcf41843e5f3510c82

This works if I integrate with respects to y but it does not work with respects to x even though I get the equation in the right form. radius = 16-(16-x^2) height = (4-x)

$$\int_{0}^{4} x^2*(4-x) dx$$

does not work, but $$V = 2 \pi \int_0^{16}(16- y)(4 - x) \, dy, x = \sqrt{16 - y} = 204.8\pi $$ works

Why is is that? I'm putting in the same radius and the height values for the cylinder respectively.

Answer 1

your integrand for the disk method is incorrect. i am not sure what the $x$ and $4-x$ represent there.

for the disk method you have $$V = \pi \int_0^4 (16 - y)^2 \, dx, y = 16 - x^2 = 204.8\pi$$ and for the shell method $$V = 2 \pi \int_0^{16}(16- y)(4 - x) \, dy, x = \sqrt{16 - y} = 204.8\pi $$