Mass Evaluation with Triple Integral

Question: Find the total mass of a solid which is located between the surface $$z=4-x^2-y^2$$ and the cone $$z=3\sqrt{x^2+y^2}$$ in the first octant, provided that its density is $$rou(x, y, z)=2$$ at all points.

This is one of the questions that I got in the test a moment ago and I'm quite curious how this problem should be solved in the right way with spherical or cylindrical coordinates. I've done numerous problems with cones bonded by a perfect sphere but have never seen a problem like this where the cone is bonded by a shape with a non-constant radius (starting from the origin in this case). Generally for sphere bonded cone problems, the $$radius$$ in cylindrical and $$rou$$ in spherical are constants, and I couldn't figure out what the corresponding $$radius$$ or $$rou$$ is in this problem.

I eventually ran out of time on solving the problem with spherical coordinates and had to use the hardcore cartesian coordinates: $$m=\int_{0} ^{1}\int _{0} ^{\sqrt{1-x^2}} \int _{3\sqrt{x^2+y^2}} ^{4-x^2-y^2} 2\,dz\,dy\,dx$$

The equation should be right but I wasn't able to evaluate this afterward as expected.

Could someone please show me how this is done in spherical or cylindrical coordinates? Thanks in advance!

• With cylindrical coordinates, $r^2=x^2+y^2$. Commented Oct 26, 2020 at 21:38
• sketch the two equations in the $x,z$ plane, then it becomes quite clear what you have to do Commented Oct 26, 2020 at 21:49
• If my answer was satisfactory, consider accepting it by clicking the tick mark button next to it @Ekidona. Commented Oct 26, 2020 at 23:43

In cylindrical, for particular $$(r,\phi),z$$ ranges from its values on the surface $$z=3r$$ to its value on the surface $$z=4-r^2$$. These two surfaces intersect at $$r=1,z=3$$. Thus, the projection of the volume on the $$(r,\phi)$$ plane is $$r\le1$$, giving

$$M=\int_0^{2\pi}\int_0^1\int_{3r}^{4-r^2}2dz~dr~d\phi=26\pi/3$$

In spherical, the two curves are $$r\cos\theta=4-r^2\sin^2\theta\implies r=\frac{\sqrt{15\sin^2\theta+1}-\cos\theta}{2\sin^2\theta}$$ and $$r\cos\theta=3r\sin\theta\implies\tan\theta=1/3$$. For a particular $$\theta,\phi,r$$ varies from $$0$$ to its value on the first surface i.e. $$\frac{\sqrt{15\sin^2\theta+1}-\cos\theta}{2\sin^2\theta}$$. For a particular $$\phi,\theta$$ varies from $$0\to\tan^{-1}(1/3)$$, giving

$$M=\int_0^{2\pi}\int_0^{\tan^{-1}(1/3)}\int_0^{\frac{\sqrt{15\sin^2\theta+1}-\cos\theta}{2\sin^2\theta}}2dr~d\theta~d\phi$$

Edit: I realized that we were only supposed to calculate the mass in the first octant. Since the mass density and volume are symmetric in the upper four octants, we can achieve the answer by dividing $$M$$ by $$4$$ without disturbing the integrals. Otherwise simply change the limit of $$\phi$$ to $$0\to2\pi/4$$ in both integrals.

• @Ekidona You substitute $x=r\sin\theta\cos\phi,y=r\sin\theta\sin\phi,z=r\cos\theta$ in both equations. Commented Oct 28, 2020 at 4:43