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I've been struggling to solve this problem of Calculus III, could you help me? For starters I have to parametrize the curve of intersection of 2 surfaces. The surfaces are:

$x² + (y-2)² = 4$

$x²+y²+z² = 16$

I need to, also, calculate the equation of the line on the point $(2,2,2√2)$.

Next, I need to determine the parameterization of the intersection of these two curves, replacing z in the first equation.

$x² + y² + z² = 1$

$z = 2(x-y)$

Now, using the spherical coordinates I need to determine the parameterization of the intersection of the curve:

$x² + y² + z² = 1$

$x = y$

Maybe I'm asking too much but it's a bit urgent!!! I've been searching all over the internet and nothing seemed to help. Thank you!

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    $\begingroup$ Please add what you have attempted, and where you are stuck. $\endgroup$ Nov 5, 2015 at 19:55

1 Answer 1

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In spherical coordinates,

$$(x, y,z)=4\,(\cos(u)\sin(v),\sin(u)\sin(v),\cos(v))$$ satisfies the equation of the sphere, as you can check by direct substitution.

Then, plugging in the equation of the cylinder,

$$x^2+y^2-4y=0=16\sin^2(v)-16\sin(u)\sin(v)=16\sin(v)(\sin(v)-\sin(u)).$$

This gives us two solutions:

  • with $\sin(v)=0, (x,y,z)=(0,0,\pm4)$;

  • with $\sin(v)=\sin(u)$, $(x,y,z)=4\,\left(\cos(u)\sin(u),\sin(u)\sin(u),\pm\cos(u)\right).$

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