we have spheres with radii of 1 and 3, and so converting to spherical we have
upon evaluating this (I'll leave that for you to do) you should get .
if you need more clarification just let me know.
I am having trouble setting up the following integral problem: "Find the volume of the solid that lies inside the cone z^2 = 3x^2 + 3y^2 and between the spheres x^2 + y^2 + z^2 = 1 and x^2 +y^2 + z^2 = 9 "
I know that it must be done with spherical coordinates and that it will be a triple integral. However, I'm a little unsure how to set up the limits and the integral itself. Thanks for any help
If I have done this right this time, hit-and-miss Monte-Carlo gives this
integral is ~=1.83 +/- ~0.02 (that's +/- 2 sigma error), or about 6-7%
of the volume of the 3x3x3 circumscribed cube.
Aparently it's gone wrong - I'll have to check the code when I get the chance.
OK now its fixed:
Integral ~=14.6+/- ~0.34 (that's +/- 2 sigma error)Code:> >NN=100000; >x=(random(NN,3)-0.5)*6; > > >nx=x(:,1)^2+x(:,2)^2+x(:,3)^2; >mx=3*x(:,1)^2+3*x(:,2)^2-x(:,3)^2; > >ll=(nx>1)&&(nx<9)&&(mx<0);ll=sum(ll')/NN*(6^3) 14.6232
RonL
I agree with your limits.
I would just multiply by 2 becaus there are two parts the upper and lower part when you draw these quadradic surfaces.
How did you get that angle for the cone? I got it from a dot product between the z-axis and the vector on a cone. How did you?
In spherical coordinates the equation,
describes a cone (half cone).
So, describes a cone above the xy-plane with angle (with z-axis) of . If we get the xy-plane. If we get the same cone but upside down. Note, these equations only describe a half-cone. A two sided cone appears in this example which is why we need to double the integral.
The question is the upper side of the cone is what equation in spherical coordinates, that is . To find the angle we need to find the angle with the lateral side of the cone (which is a straight line) and the z-axis (that is definition of ). To do that we can use the dot product. Select a point on the upper-cone, say . Select a point on the z-axis .
The vector from the origon to point on cone is and the vector from origon to z-axis is .
Now you take the dot product to find the angle,
Which is what you get.