Find Volume b/w 2 Surfaces: x, y, z Equations

  • Thread starter Thread starter Kuma
  • Start date Start date
  • Tags Tags
    Volume
Kuma
Messages
129
Reaction score
0

Homework Statement



Find the volume between the two surfaces

z = 2x^2 + y^2

z = 4 - y^2

Homework Equations





The Attempt at a Solution



Ok so i found out that the surfaces intersect at a circle. When i solved i got

x^2 + y^2 = 2, so the circle has a radius of sqrt 2.

So these are the bounds i got. X goes from - sqrt 2 to sqrt 2, y goes from - sqrt 2-x^2 to sqrt 2-x^2 and z goes from 4-y^2 to 2x^2 + y^2.

Is that right? If so, what is the equation I am integrating? Is it always integrating 1 dx dy dz?
 
Physics news on Phys.org
That would certainly work. And what would be the bounds on the z-integral?


(Also, because of the circular symmetry, I would be inclined to use cylindrical coordinates- but you certainly can do it the way you suggest.)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top