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I have this weird solid to evaluate which is the solid constrained under the sphere : $x^2+y^2+(z-3)^2=9$ and over the bottom half cone described as $z=2-\sqrt{x^2+y^2}$. and I have to evalute the triple integral (x^2+y^2) dV...

So I'm being confused by this problem as the result I get between the two coordinates are not the same but they are slightly close. Working in spherical, I found:

  • arctan(((0.438)/(1.56)))< phi < pi/2
  • 0 < theta < 2*pi
  • 6*cos(phi) < rho < 2/(cos(phi)+sin(phi)

Finding ±385 cubic unit

Working in cylindrical, I found, (Solid one)

  • 2-r < r < √(9-r^(2))+3
  • 0 < z < ((√(17)-1)/(2))
  • 0 < theta < 2pi

(Solid two)

  • (√(9-r^(2))-3) < r < √(9-r^(2))+3
  • ((√(17)-1)/(2)) < z < 3
  • 0 < theta < 2pi

Finding ±400 cubic unit

My question was defined as off topic? I don't know what more I can say, I just don't see why the result are different, I thought I was supposed to find the same element of volume no matter which system I use. I can't see where I'm making my mistake.

enter image description here

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  • $\begingroup$ I think you mean "cone" not "cylinder" $\endgroup$ – Hugh Jul 17 '17 at 22:46
  • $\begingroup$ Have you tried using cylindrical coordinates? The problem is symmetrical around the z axis so this is the most natural coordinate system. $\endgroup$ – Hugh Jul 17 '17 at 22:47
  • $\begingroup$ Using cylindrical I have come with a solution with two E element of solid and two D domain of integration? $\endgroup$ – Anthony Charest Jul 17 '17 at 22:51
  • $\begingroup$ By "under the sphere", do you mean "under the top hemisphere" or do you mean "within the sphere"? $\endgroup$ – Robert Israel Jul 18 '17 at 0:20
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Update: Full solution.

Call the solid in question $B$. We are told to compute its moment of inertia with respect to the $z$-axis, i.e. the integral $$\Theta:=\int_B(x^2+y^2)\>{\rm d}(x,y,z)\ .$$ The problem is rotationally symmetric with respect to the $z$-axis. We therefore shall use cylindrical coordinates. Put $\rho:=\sqrt{x^2+y^2}\geq0$ and draw a figure in the $(\rho,z)$ meridian half plane. In this figure we see the right half of the circle $\rho^2+(z-3)^2=9$ and the half line $z=2-\rho$. This half line intersects the circle at a point $(\rho_*, z_*)$ found by solving$$\rho+z=2,\qquad \rho^2=6z-z^2\ .$$ One obtains $z_*={1\over2}\bigl(5-\sqrt{17}\bigr)$.

Horizontal planes $z={\rm const.}$ with $z_*\leq z\leq2$ intersect the solid $B$ in an annulus of inner radius $2-z$ and outer radius $\sqrt{6z-z^2}$, and when $2\leq z\leq 6$ we obtain a circular disc of radius $\sqrt{6z-z^2}$. In cylindrical coordinates the volume element is given by ${\rm d}V=2\pi\rho\> d\rho\>dz$. We therefore obtain $$\eqalign{\Theta&=2\pi\int_{z_*}^2\int_{2-z}^{\sqrt{6z-z^2}} \rho^2\cdot\rho\>d\rho\>dz+2\pi\int_2^6\int_0^{\sqrt{6z-z^2}}\rho^2\cdot\rho\>d\rho\>dz\cr &={17\pi\over20}\bigl(89+15\sqrt{17}\,\bigr)\doteq402.814\ .\cr}$$ In any case I would not advise spherical coordinates for this problem, whether centered at $(0,0,0)$, $(0,0,2)$, or $(0,0,3)$, since the apex of the cone and the center of the sphere do not coincide. Note that, apart from the factor $\pi$, the end result is an algebraic number that you would not have been able to identify using the geographical latitude $\theta$, trig and inverse trig functions.

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  • $\begingroup$ I understand this very well, I indeed had this result using the cylindrical coordinate system but I failed to obtain it in a spherical coordinate, and I thought that both system were supose to send the exam same volum :S Thank you for that clarification sir! $\endgroup$ – Anthony Charest Jul 18 '17 at 16:10
  • $\begingroup$ I think I forgot to mention the function...which is (x^2+y^2)..... $\endgroup$ – Anthony Charest Jul 18 '17 at 16:59

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