Anonymous
Anonymous asked in 科學數學 · 1 decade ago

高等微積分~~

Revolve the parametrized curve c about the x-axis (a)Show that r(u,v)=f(u)i+g(u)cosvj+(g(u)sinv)k

(b)Find the parametrization of revolving the curve x=y^2,y>=0 about the x-axis.

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  • 1 decade ago
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    (a) Without loss of generality the parametrized curve c is C: x=f(u), y=g(u), where u is the parameter. When a plan curve is revolved about the x-axis, each point (x,y) will generate a circle of radius |y|, in a second parameter v , (x,y) will generate the circle (x,ycosv,ysinv), v=0 to 2pi. In this regard the curve C would generate a surface, using two paremeters u and v: S: x=f(u), y=g(u)cosv, and z=g(u)sinv, v=0 to 2pi. This is equivalent to the desired expression r(u,v)=f(u)i+g(u)cosvj+(g(u)sinv)k, v=0 to 2pi

    (b)We may parametrize x=y^2 using x=f(u)=u^2, and y=g(u)=u, u=0 and more. Make use of part (a) then r(u,v)=u^2i+ucosvj+usinvk, u greater than or equal to 0.

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