how would you do the triple integral pf z(x+y) where E is the solid bounded by?
the cylinders x^2+y^2=1 and x^2+y^2=4, the paraboloid z=x^2+y^2
the x-y plane and to the left of the y-z plane (solid is in the 2ndand 3rdoctants!)
- RealProLv 76 months agoBest Answer
Octants are stone age. Just say x<0?
Before starting, you make the argument that the entire region is symmetric about xz plane and the integrand is an odd function of y, so you can just remove the +y part. The integral on the part of the region where y>0 cancels out with the one where y<0. You can try it without removing anything but it's extra work.
You can try cartesian for fun. If your outermost integral is along y, then the solution is the sum of 3 integrals, and if it's along x it's a sum of 2.
Preferred choice of action is cylinderical under the standard notation since bounds call for it and integrand is easily transformed.
Let the outermost integral be along φ, and inside each step we do it along r and inside that z.
∫ [pi/2 to 3pi/2] ∫ [1 to 2] ∫ [0 to r^2] z(r cosφ + r sinφ) r dz dr dφ
Where the r at the end is due to the geometry of the transformation.
Now ditch the y term
∫ [pi/2 to 3pi/2] ∫ [1 to 2] ∫ [0 to r^2] zr^2 cosφ dz dr dφ
∫ [pi/2 to 3pi/2] ∫ [1 to 2] (1/2)r^6 cosφ dr dφ
∫ [pi/2 to 3pi/2]cosφ dφ ∫ [1 to 2] (1/2)r^6 dr
-2 * (64 - 1/2)/7
- az_lenderLv 76 months ago
Those three surfaces do not enclose a bounded volume. The two cylinders are concentric, but the paraboloid cuts each of them only along a circle. You need a fourth surface (the xy-plane, for example), to make a closed volume.