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Molodets [167]
2 years ago
13

A right circular cone is intersected by a plane that passes through the cone's vertex and along the edge of each nappe, what is

produced from this intersection?
Mathematics
1 answer:
Vsevolod [243]2 years ago
5 0
<span>This intersection produces a line. However, if the angle of the plane is less than the cone's angle, then the intersection produces a point. If the angle of the plane is greater than the angle of the cone, then the intersection is two lines intersecting at the vertex. If the plane intersects at some point other than the vertex, then the intersection is a circle if the plane is perpendicular to the cone's axis. It is an ellipse when the plane's angle is less than the cone's angle. It's a parabola when the planes's angle equals the cone's angle.</span>
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Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

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\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
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\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
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Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
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\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
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Step-by-step explanation:

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