[無料ダウンロード！ √] cone z=sqrt(x^2 y^2) 268431

 how do i plot the section of a cone z = 9sqrt(x^2 y^2) in the cylinder of r=2 Follow 2 views (last 30 days) Show older comments Carlos Perez on Vote 1 ⋮ Vote 1 Commented John D'Errico on pretty much what the question says ive tried two different ways and none of them have worked I can post what i have if

Cone z=sqrt(x^2 y^2)-Example 1 Parametrize the single cone $z=\sqrt{x^2y^2}$ Solution For a fixed $z$, the cross section is a circle with radius $z$So, if $z=\spfv$, the Notice that the bottom half of the sphere `z=sqrt(1(x^2y^2))` is irrelevant here because it does not intersect with the cone The following condition is

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Solution for z=sqrt(x^2y^2)fory equation Simplifying z = sqrt(x 2 y 2) * fory Reorder the terms for easier multiplication z = qrst * fory(x 2 y 2) Multiply qrst * fory z = foqr 2 sty(x 2 y 2) z = (x 2 * foqr 2 sty y 2 * foqr 2 sty) z = (foqr 2 stx 2 y foqr 2 sty 3) Solving z = foqr 2 stx 2 y foqr 2 sty 3 Solving for variable 'z' Move all terms containing z to the left, all \\begin{array}{c} 1 \le y \le 1\\ 0 \le x \le \sqrt {1 {y^2}} \\ {x^2} {y^2} \le z \le \sqrt {{x^2} {y^2}} \end{array}\ The first two inequalities define the region \(D\) and since the upper and lower bounds for the \(x\)'s are \(x = \sqrt {1 {y^2}} \) and \(x = 0\) we know that we've got at least part of the right half a circle

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