- #1
KStolen
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I'm taking a Calculus class as an elective. This might not have been a good idea, but I'm stuck in it now.
Here is a problem I have to do. My knowledge of basic maths is poor, so please be gentle and explain thoroughly!
1. The problem.
Rewrite the following integral in terms of spherical polar coordinates.
[tex]\int\int\int x^{2}z^{2}exp((x^{2}+y^{2}+z^{2})/a^{2})dxdydz[/tex]
Now evaluate it over the region bounded by the planes [tex]x = 0, y = 0, z = 0[/tex]
and the sphere [tex]x^{2} +y^{2} +z^{2} = a^{2}[/tex] in the first octant.
Spherical co-ordinates [tex](r,\vartheta,\phi)[/tex] to rectangular co-ordinates [tex](x,y,z)[/tex] are as follows :
[tex]x=rSin\vartheta Cos\phi[/tex]
[tex]y=rSin\vartheta Sin\phi[/tex]
[tex]z=rCos\vartheta[/tex]
[tex]dV=dxdydz=r^{2}Sin\vartheta dr d\vartheta d\phi[/tex]
[tex]x^{2}+y^{2}+z^{2} = r^{2}Sin^{2}\vartheta Cos^{2}\phi + r^{2}Sin^{2}\vartheta Sin^{2}\phi + r^{2}Cos^{2}\vartheta[/tex]
[tex]x^{2}+y^{2}+z^{2} = r^{2}(Sin^{2}\vartheta + Cos^{2}\vartheta)[/tex]
[tex]x^{2}+y^{2}+z^{2} = r^{2}[/tex]
But [tex]a^{2}[/tex] is also [tex]x^{2}+y^{2}+z^{2}[/tex]
So the integral evaluates to
[tex]\int\int\int r^{2}Sin^{2} \vartheta Cos^{2}\phi r^{2}Cos^{2}\vartheta exp(a^{2}/a^{2})r^{2}Sin\vartheta dr d\vartheta d\phi[/tex]
Which evaluates to
[tex]e\int\int\int r^{6}Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi dr d\vartheta d\phi[/tex]
So I've written it in spherical co-ordinates. I think everything is right so far. (hopefully I don't have any typos in my LaTeX code)
The next part is what I'm unsure about. I don't know how to set up the boundaries.
No doubt this is trivial, but it's also a central part of doing the integral.
Here's an attempt :
[tex]e\int^{\pi}_{0}\int^{\pi/2}_{0}\int^{a}_{0} r^{6}Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi dr d\vartheta d\phi[/tex]
Could somebody explain to me exactly what I need to do to find the boundaries?
Thanks!
Here is a problem I have to do. My knowledge of basic maths is poor, so please be gentle and explain thoroughly!
1. The problem.
Rewrite the following integral in terms of spherical polar coordinates.
[tex]\int\int\int x^{2}z^{2}exp((x^{2}+y^{2}+z^{2})/a^{2})dxdydz[/tex]
Now evaluate it over the region bounded by the planes [tex]x = 0, y = 0, z = 0[/tex]
and the sphere [tex]x^{2} +y^{2} +z^{2} = a^{2}[/tex] in the first octant.
Homework Equations
Spherical co-ordinates [tex](r,\vartheta,\phi)[/tex] to rectangular co-ordinates [tex](x,y,z)[/tex] are as follows :
[tex]x=rSin\vartheta Cos\phi[/tex]
[tex]y=rSin\vartheta Sin\phi[/tex]
[tex]z=rCos\vartheta[/tex]
[tex]dV=dxdydz=r^{2}Sin\vartheta dr d\vartheta d\phi[/tex]
The Attempt at a Solution
[tex]x^{2}+y^{2}+z^{2} = r^{2}Sin^{2}\vartheta Cos^{2}\phi + r^{2}Sin^{2}\vartheta Sin^{2}\phi + r^{2}Cos^{2}\vartheta[/tex]
[tex]x^{2}+y^{2}+z^{2} = r^{2}(Sin^{2}\vartheta + Cos^{2}\vartheta)[/tex]
[tex]x^{2}+y^{2}+z^{2} = r^{2}[/tex]
But [tex]a^{2}[/tex] is also [tex]x^{2}+y^{2}+z^{2}[/tex]
So the integral evaluates to
[tex]\int\int\int r^{2}Sin^{2} \vartheta Cos^{2}\phi r^{2}Cos^{2}\vartheta exp(a^{2}/a^{2})r^{2}Sin\vartheta dr d\vartheta d\phi[/tex]
Which evaluates to
[tex]e\int\int\int r^{6}Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi dr d\vartheta d\phi[/tex]
So I've written it in spherical co-ordinates. I think everything is right so far. (hopefully I don't have any typos in my LaTeX code)
The next part is what I'm unsure about. I don't know how to set up the boundaries.
No doubt this is trivial, but it's also a central part of doing the integral.
Here's an attempt :
[tex]e\int^{\pi}_{0}\int^{\pi/2}_{0}\int^{a}_{0} r^{6}Sin^{3}\vartheta Cos^{2}\vartheta Cos^{2}\phi dr d\vartheta d\phi[/tex]
Could somebody explain to me exactly what I need to do to find the boundaries?
Thanks!