Triple Integral (Calc III-Multivariable Calc)

[MhailJ]

Homework Statement

Integrate the function f(x,y,z)=3x+8y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y=(sqrt(17/47))*x and contained in a sphere centered at the origin with radius 10 and a cone opening upwards from the origin with top radius 8.

Homework Equations

I know that we will use both the planes, x=0 and y=sqrt17/47. As well as the sphere (x^2+y^2+z^2=10) and the cone (x^2+y^2=8z^2).

The Attempt at a Solution

I understand that this shape will look something like an ice cream cone, with a plan in the yz-plane and the other given plane as bounds. However, I do not understand what bounds go where. Due to the fact that it is a "slice" of an ice cream cone, does this mean spherical coordinates must be used to solve this equation? I do not quite understand what solid they want to integrate with respect to. Does this mean that the solid is IN the cone? If so, why do they give me the sphere that the cone is in- because the top of the cone ends before the sphere ends, therefore giving the "ice cream cone" type look to this figure? If someone could just give me a start with this problem in terms of what it looks like I think I can work out the rest...I know this is quite difficult to show online due to the fact that I cannot show you guys my sketch that I currently have done.

Thank you very much!
MhailJ

 

[jackmell]

It looks barely like some quadrant of the cone. Also, if the radius of the sphere is 10, then that would be x^2+y^2+z^2=100 right? But it won't qualitatively change the problem. Here's what I suggest: It's good to try and draw them free-hand . . . for a little while, then learn how to do them in a CAS like Mathematica. The plot below is what I did so I assume it's over that green piece in there. Red is the cone, blue is the paraboloid, green the sphere and dark-green x=0. Alright, assume that's what you want. Now, just for starters, do it in cartesian coordinates even though that will probably be too complex to integrate but that at least will help you visualize how to integrate it. So, consider:

$$\int\int\int f(x,y,z) dzdydx$$

so at least conceptually, the dz is easy right? It's just in that green piece starting from the cone, up to the sphere:

$$\int\int\int_{red}^{light green} f(x,y,z) dzdydx$$

I'll do the red one for you. If it's equation is [itex]x^2+y^2=8z^2[/tex], then my lower bound on z would be:<br /> <br /> $$\sqrt{1/8(x^2+y^2)}$$<br /> <br /> You do the other one. Now what about y? Isn't that going from the border of the blue paraboloid to the border where the cone and sphere intersect? So the lower bound on y would be just the equation of the paraboloid in the x-y plane. You try and figure where the equations [itex]x^2+y^2+z^2=10[/tex] and [itex]x^2+y^2=8z^2[/tex] interesect for the top limit. So far then we have:<br /> <br /> $$\int_a^b \int_{\sqrt{17/47 x}}^{myintersection}\int_{1/8(x^2+y^2)}^{mysphere} f(x,y,z)dzdydx$$<br /> <br /> Now you try and finish it.[/itex][/itex][/itex]