Question

just set up a triple integral in rectangular cordinates to compute the volume of ellipsoid

x^2/a^2+y^2/b^2 - z^2/c^2 = 1, a b and c are real constants, specify the order of integration and find the limits of integration.

Answer 1

x^2/a^2+y^2/b^2 - z^2/c^2 = 1 is not an ellipsoid, its a hyperboloid with one sheet.
x^2/a^2+y^2/b^2+z^2/c^2=1 is...
ok so the region of this in R^2 is :

where
y_1(x)=b*sqrt(1-x^2/a^2)
y_2(x)=-b*sqrt(1-x^2/a^2)
now our x as you can see is just between -a/2 and a/2. so we'll have limits:
-b*sqrt(1-x^2/a^2)<=y<=b*sqrt(1-x^2/a^2)
-a/2<=x<=a/2
now in the 3D:

where
z_1(x,y)=c*sqrt(1-x^2/a^2-y^2/b^2)
z_2(x,y)=-c*sqrt(1-x^2/a^2-y^2/b^2)
so our limits for z is:
-c*sqrt(1-x^2/a^2-y^2/b^2)<=z<=c*sqrt(1-x^2/a^2-y^2/b^2)
so we'll have the triple integral:
\[\int\limits_{-a/2}^{a^2} \int\limits_{-(b \sqrt{1-x^2/a^2})}^{(b \sqrt{1-x^2/a^2})}\int\limits_{-c \sqrt{1-x^2/a^2-y^2/b^2}}^{c \sqrt{1-x^2/a^2-y^2/b^2}}dV\]