Recent content by MorallyObtuse

Hi, Set up the triple integral in spherical coordinates to find the volume bounded by $$z = \sqrt{4-x^2-y^2}$$, $$z=\sqrt{1-x^2-y^2}$$, where $$x \ge 0$$ and $$y \ge 0$$. $$\int_0^{2\pi} \int_0^2 \int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}} r\ dz\ dr\ d\theta$$

Polar Inertial Momentum Inequality

That's true, maybe I'm a little too ungrateful. Put it this way, I'm not the fastest learner.

Forget it! I barely understand whenever you help me.

Yeah, I agree with you, Mark :)

The teacher uses 'baby talk notation' like +ve

The questions are close. So, not much difference in the answers. You might want to simply prove that "if a> b and k< 0 then ka< kb" first. Then prove (a) by letting x= a, y= b, and prove (b) by letting x= b, y= a. This part I'm not understanding. I'd have to input values and the teacher says...

a.) Since x > y, so x - y is positive and k is negative. Product of a negative and positive number is negative, kx - ky Hence it follows that kx < ky. b.) Since x < y, so x - y is negative and k is negative. Product of two negative numbers is equal to a positive number. Hence it follows...

Hi, Are these correct? Homework Statement a.) Given that x > y, and k < 0 for the real numbers x, yand , show that kx < ky. b.) Show that if x, y ∈ R, and x < y , then for any real number k < 0,kx > ky 2. The attempt at a solution a.) kx > y...1 x > y x - y is +ve...2 k...

Ok, thanks!

Homework Statement Express x^4 + y^4 + z^4 - 2y^2z^2 - 2z^2x^2 - 2x^2y^2 as the product of four factors. 2. The attempt at a solution (x-y-z)(-x+y-z)(-x-y+z)(x+y+z)

And thanks very much :) (a-b) (a^2+ab+b^2) (a+b)(a^2-ab+b^2)

I did get it, just joking with 1/2" :biggrin:

No I don't get it:biggrin::biggrin::biggrin:

I don't get it?!