Spherical coordinates triple integral

brad sue
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Hi,

Please can someone help me with this problem:

find the triple integral over T( using spherical coordinate)

T: 0<=x<=1
0<= y<=sqrt(1-x^2)
sqrt(x^2+y^2)<= z <= sqrt(2-(X^2+y^2))


please help me just to find the boundaries of the integrals.
I tried but I did not find the solution of the textbook. ( because I set the wrong triple integral) I also tried to draw a picture but ...nothing


Thank you
 
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Use the relations between Cartesian (x,y,z) and spherical coordinates (r,\theta,\phi[/tex]) to substitute for x, y and z:<br /> <br /> x=rsin(\phi)cos(\theta)<br /> y=rsin(\phi)sin(\theta)<br /> z=rcos(\phi)<br /> <br /> where phi is the angle between a vector and the z-axis. theta is the angle between the projection on the x,y plane ad the x-axis.
 
Also don't forget your Jacobian, in this case being r²sin(phi).
 
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...
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