Recent content by planck42

Oh duh the cone isn't a bound on rho. In that case, I agree with the integral you've got and arrived at an answer of \frac{1}{60}(8\sqrt{2}-10), which is approximately 0.022

Why is the lower bound for rho zero? The sphere meets the cone well before that.

I don't quite understand what you're trying to say; are you referring to the bounds in the translated or to those in the untranslated coordinate system? If the former, then I agree [that the lower bound is 0] but the upper bound is no longer sqrt 2, and it can't be the latter.

What does that have to do with anything? If you're implying that rho be from 0 to 2, then that's wrong because rho from 0 to two means the entire region under the sphere of radius 2. By the way, the answer I got was this...

I just got a better idea: when the cone hits the cylinder, z=r=1, but conveniently z=1 is where the cone intersects the sphere! Therefore, a dome-like shape is the region of integration, and if one were to move the xy-plane up to the circle which is the base of the dome, then it would get rid of...

First off, the restrictions on x and y make it clear that theta goes from 0 to pi/2. Next, phi goes from 0 on the sphere which was pointed out by SammyS(I didn't notice that initially, but it's a tremendous boost) to pi/4 on the cone. That leaves rho, which ends at sqrt 2 and begins... where...

Homework Statement Evaluate by changing to spherical coordinates \int^1_0\int^{\sqrt{1-x^2}}_0\int^{\sqrt{2-x^{2}-y^2}}_{\sqrt{x^{2}+y^2}}xydzdydx Homework Equations dz dy dz = {\rho}^{2}sin{\phi} The Attempt at a Solution This problem is quite simple to do in cylindrical...

0 to 4-2x-2y. But how did you get y being from 0 to 2-x for fixed x only?

Let's just suppose that x is going from 0 to 2(it should be equivalent to taking z from 0 to 4 if I do this right). Then for a fixed value of x, y should be \frac{4-z-2x}{2}. For a fixed x and y, z should be 4-2x-2y. The problem I have with this is that if I find z for fixed x, I get a z that...

I don't find it possible to answer both questions as there is only one boundary function.

That makes sense; but now I feel even more lost.

Homework Statement {\int}{\int}{\int}ydV over the region E, where E is bounded by x=0, y=0, z=0, and 2x+2y+z=4 Homework Equations n/a The Attempt at a Solution Assuming that x and y must both be positive, which the boundary conditions seem to require, then the most either one can...

I believe that it is possible for free electrons to flow through the battery itself without the need for a wire, but it wouldn't be desirable.

Infinity is much more of a mathematical concept than anything real, but periodicity is a good way to disguise it. For instance, closed paths are essentially periodic, and it's possible to go an infinite distance along them with a finite amount of space.

Open-ended questions in science are the doom of you? What kind of open-ended questions?