calculus - find the equation of the cone $z=\sqrt{x^2+y^2
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Jul 19, 2013 · On the top horizontal plane where the cone cuts the sphere, z = 1/sqrt (2) and the radius R of the circular cross-section is given by R = sqrt (x^2+y^2) = z = 1/sqrt (2), so that phi = pi/4 for the
`z=+-sqrt(1-(x^2+y^2))` Notice that the bottom half of the sphere `z=-sqrt(1-(x^2+y^2))` is irrelevant here because it does not intersect with the cone. The following condition is true to find the
Find the volume of the solid that is enclosed by the cone z = sqrt (x^2 + y^2) and the sphere x^2 + y^2 + z^2 = 2
Nov 10, 2020 · The lower bound \(z = \sqrt{x^2 + y^2}\) is the upper half of a cone and the upper bound \(z = \sqrt{18 - x^2 - y^2}\) is the upper half of a sphere. Therefore, we have \(0 \leq \rho \leq \sqrt{18}\), which is \(0 \leq \rho \leq 3\sqrt{2}\)
Find a parametric representation for that part of the sphere of radius {eq}12 {/eq} centered at the origin that lies above the cone {eq}z =\sqrt{x^2 + y^2} {/eq}. Parametric Representation of a Curve:
To find the spherical coordinate equation for the cone : $$z = \sqrt {x^2 + y^2} $$ We have the transformation given by : {eq}x=\rho \cos \theta \sin \phi {/eq}
Sep 26, 2017 · karush said: \tiny {15.4.17} Find the volume of the given solid region. bounded below by the cone. z=\sqrt {x^2+y^2} and bounded above by the sphere. x^2+y^2+z^2=128. using triple integrals. \displaystyle\int_ {a}^ {b}\int_ {c}^ {d} \int_ {e}^ {f} \,dx\,dy \,dz
Question 1011000: Find the surface area of the cone z=sqrt(x^2+y^2) below the plane z=8. Please show your solution step by step. Answer by rothauserc(4717) (Show Source): You can put this solution on YOUR website! We want the surface area of the portion of the cone z^2 = x^2 + y^2 between z=0 and z=8. The equation of the cone in cylindrical
Apr 16, 2013 · Find an equation for the cone z=sqrt(x^2+y^2) in spherical coordinates.? ϕ = ? PLEASE HELP! Thank you! Answer Save. 2 Answers. Relevance. Hey You. Lv 4. 8 years ago. Favorite Answer. z in spherical: r Sin phi. x^2+y^2 = r^2 Cos^2 phi. r Sin Phi = r Cos Phi. Tan Phi = 1. Phi = pi/4. 0 0. Karen. 8 years ago. 88. Source(s): school. 0 1. Still
Dec 20, 2012 · x^2 + y^2 + z^2 = 9. x^2 + y^2 + (x^2 + y^2) = 9. 2x^2 + 2y^2 = 9. x^2 + y^2 = 4.5. This is a circle with radius sqrt(4.5) <---from the equation for the sphere. That is at z = 4.5 <---from the equation for the cone. It looks sort of like an ice cream cone, a cone with a bit of stuff on top that is formed by the sphere defined by the circle x^2
Find the volume of the solid which is above the cone z=sqrt(x^2+y^2) and inside the sphere given by. x^2+y^2+z^2=18.. Hint: Solve for the curve which is the intersection of these two geometric surfaces.(This is a calc 3 probelm involving double or triple integrals)(please hurry and answer, this problem is due at 12:00a.m)
Jul 19, 2013 · On the top horizontal plane where the cone cuts the sphere, z = 1/sqrt (2) and the radius R of the circular cross-section is given by R = sqrt (x^2+y^2) = z = 1/sqrt (2), so that phi = pi/4 for the
Find the spherical coordinate equation for the cone {eq}z = \sqrt {x^2 + y^2} {/eq}. Spherical Coordinates : In space, the locus can be expressed in cartesian or Spherical or Cylindrical form of
A solid lies above the cone z = \sqrt{x^2 + y^2} and below the sphere x^2 + y^2 + z^2 = z . Write a description of the solid in terms of inequalities involving…
Sep 26, 2017 · $\tiny{15.4.17}$ Find the volume of the given solid region bounded below by the cone $z=\sqrt{x^2+y^2}$ and bounded above by the sphere $x^2+y^2+z^2=128$
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