Double integrals?

Hey can anyone help me with my double integrals homework? I have to use double integrals to find volumes...my question is:





Determine the volume of the solid that lies inside of the ellipsoid: x^2/a^2+y^2/a^2+z^2/c^2=1, above the xy plane and inside of the cylinder x^2+y^2-ay=0.





Thanks for your help!

Double integrals?
This is actually a triple integral.


The cylinder has radius a/2 and its central axis is y = a/2.


Converting to polar cordinates


ellipse: r^2 / a^2 = 1 - z^2 / c^2


cylinder: r = a*sinθ


They intersect when sin^2 θ = 1 - z^2 / c^2 or


cos^2 θ = z^2 / c^2


cosθ = +/- z/c





So these are your limits of integration:


For 0 %26lt; θ %26lt; arccos(z/c), 0 %26lt; r^2 %26lt; a^2 * sin^2 θ


For arccos(z/c) %26lt; θ %26lt; π/2, 0 %26lt; r^2 %26lt; 1 - z^2 / c^2


0 %26lt; z %26lt; c





Since the figure is symmetrical about the yz plane, we only need to perform the integration from 0 %26lt; θ %26lt; π/2 and then multiplly the result by 2.





So we need to integrate:


r*dr*dθ*dz between those limits


or 1/2 * d(r^2)*dθ*dz





1) integration between 0 %26lt; θ %26lt; arccos(z/c)


It's a straightforward integration just putting int he limits and simplifying to get:


a^2 / 2 * [arccos(z/c) - z/c^2 * sqrt(c^2 - z^2)] * dz





2) integration between arccos(z/c) %26lt; θ %26lt; π/2


Again after simplification


(a^2 / c^2) * (π/2 - arccos(z/c)) * (c^2 - z^2) * dz





3) So we need to integrate those two between 0 %26lt; z %26lt; c, add them, then multiply the result by 2.


After simplification, once again, we finally need to integrate:


(1/2)*a^2 [(2*z^2/c^2 - 1)*arccos(z/c) - (z/c^2)*sqrt(c^2 - z^2) - π*z^2 / c^2 + π/2 ] * dz





Use the site below to help with some of that integration. You should get:





[a^2/(18*c^2)] * [ {sqrt(c^2-z^2)*(5c^2-2z^2) + (6z^3-9*c^2*z)*arccos(z/c) } + 3*(c^2-z^2)^(3/2) - 3*π*z^3 + 9*π*c^2*z ]





After putting in the limits,


V = a^2*c/9 * (3π - 4)
Reply:The answer is (a^2)x c x(pi/3 - 4/9). To show that, you change your coordinates to u=x/a, v=y/a and w=z/c. Then you go to polar coordinates in the (u,v) plane. The conditions become





r^2 + w^2%26lt;1 and r %26lt; sin(theta) for theta between 0 an d pi. You intergrate first in w and get





int(sqrt(1--r^2) r dr d theta)





Then you integrate in r on the intervall ( 0, sin theta), and finally in theta.
Reply:Well, I dont understand exactly what the question means, but the formula I was taught for finding volumes by intergration is:





V= (pi)*[(integral from b to a of)y^2 dx]





where y is the equation of the curve you need to find the volume of when rotated around the x axis.