Evaluate the triple integral (with spherical coordinates)

melihaltintas
Messages
6
Reaction score
0

Homework Statement


Firstly sorry for my bad english,i have a one question for you(İ try it but i didn't solve it )

Homework Equations


The Attempt at a Solution


i know problem will be solved spherical coordinates but i don't know how i get angles (interval) theta and fi ?
\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}e^{-\sqrt{x^2+y^2+z^2}}\,dxdydz

Mod note: Fixed LaTeX. Read what tiny-tim said below.[/color]
 
Last edited by a moderator:
Physics news on Phys.org


You get the angles by thinking about what the spherical coordinate variables mean. \phi is the angle from the positive z-axis to the negative x-axis and, to cover all space, would normally go from 0 to \pi. Since you want z to stay positive, you want \phi to go from 0 to \pi/2. \theta goes around a complete circle in the xy-plane in going from 0 to 2\pi. Since the first quadrant is 1/4 of that, \theta goes from 0 to \pi/2. Finally, there is no upper limit on the distance from the origin to any point in the first octant so \rho goes from 0 to \infty.
 
welcome to pf!

hi melihaltintas! welcome to pf! :smile:

don't try to put forum tags inside latex! :wink: …​
melihaltintas said:
\int_{0}^{\infty}\int_{0}^{ \infty}\int_{0}^{\infty}e^{-\sqrt{x^2+y^2+ z^2}}\,dxdydz
 
thanks a lot :)
 
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...
Back
Top