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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 coordinates is just z = r, so we can take as our parameters r and t (representing theta) ***** treat that potion(S) of the cone as a graph whose shadow D on the xyplane is the disk of radius 8 The Sphere is x² y² z² = 1 And cone z = 6√(x² y²) The volume of the Sphere is Where x,y and z are in Cartesian coordinates and Ѳ, φ and ρ are in Spherical coordinate systemLast, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form z 2 = x 2 a 2 y 2 b 2 z 2 = x 2 a 2 y 2 b 2 In this case, we could choose any of the three However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might

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Graph of cone z=sqrt(x^2 y^2)-Answer to Use the parameterization of the cone frustrum z=2 \sqrt{x^2y^2} between the planes z=4 and z=6 to express the area of theIn this video we discuss the formulas you need to be able to convert from rectangular to spherical coordinates We then convert the rectangular equation for

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 Section 15 Functions of Several Variables In this section we want to go over some of the basic ideas about functions of more than one variable First, remember that graphs of functions of two variables, z = f (x,y) z = f ( x, y) are surfaces in three dimensional space For example, here is the graph of z =2x2 2y2 −4 z = 2 x 2 2 y 2 − 4 Now, let's see what the range for \(z\) tells us The lower bound, \(z = \sqrt {{x^2} {y^2}} \), is the upper half of a cone At this point we don't need this quite yet, but we will later The upper bound, \(z = \sqrt {18 {x^2} {y^2}} \), is the upper half of the sphere, \{x^2} {y^2} {z^21 This figure is the (double) cone of equation x 2 = y 2 − z 2 The gray plane is the plane ( x, y) You can see that it is a cone noting that for any y = a the projection of the surface on the plane ( x, z) is a circumference of radius a with equation z 2 x 2 = a 2

We know how to parametrize the circle x 2 y 2 = r 2 x = r cos ⁡ θ y = r sin ⁡ θ So we'll parametrize the cone using the same method, only r will be a variable instead of a constant z = x 2 y 2 z 2 = x 2 y 2 Let z = ρ, 0 ≤ ρ ≤ 4 x 2 y 2 = ρ 2 Parametric equation x = ρ cos ⁡ θ y = ρ sin ⁡ θ z = ρ 0 ≤ ρ ≤ 4 0 ≤ θ ≤ 2 πAnswer to Given the cone, S_1, z = sqrt(x^2 y^2), and the hemisphere, S_2, z = sqrt(2 x^2 y^2);(e) Below is the graph of z = x2 y2 On the graph of the surface, sketch the traces that you found in parts (a) and (c) For problems 1213, nd an equation of the trace of the surface in the indicated plane Describe the graph of the trace 12 Surface 8x 2 y z2 = 9;

2 Let T be the solid bounded by the paraboloid z= 4 x2 y2 and below by the xyplane Find the volume of T (Hint, use polar coordinates) Answer The intersection of z= 4 2x 22y and xyplane is 0 = 4 x2 y;ie x2 y = 4 In polar coordinates, z= 4 x2 y 2is z= 4 rSo, the volume is Z Z 4 x2 y2dxdy = Z 2ˇ 0 Z 2 0 4 r2 rdrd = 2ˇ Z 2 0 4r r3 2 drThe square root keeps us from going above that point z=4 if we manipulate the equation and isolate x 2 y 2 we get x 2 y 2 = 16 z 2 (remember that since we have a square root in our original function, we have to consider it's domain in our graph, meaning zFind the surface area of the part of the cone z = sqrt(x 2 y 2) that lies between the plane y=x and the cylinder y=x 2 Expert Answer Who are the experts?

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Squared is equal to x squared, plus y squared So And if we have Z equals zero, we get zero equals X squared plus y squared This implies that X equals zero y equals zero Okay, so we have a vertex here and we have our cone shape So let's look at so if y is equal to zero z squared equals x squared so that Z equals plus or minus XX^2y^2z^2=9 graph X^2y^2z^2=9 graphFor example, the graph of x2 y2 = 4 is the cylinder obtained by translating the circle x2 y2 = 4 of radius 2 centered at the origin in the xyplane along the zaxis Similarly, the graph of the surface F (x;z) = 0 can be obtained by translating the curve 11 x2 y 2 9 12 y2 z 2= 9 13 x y2 z 4 14 1 x2 y2 z 2 4The graph of the function EXERCISE 7 f (x, y) = x 2 y 2 is the cone z = x 2 y 2 Find the equation of the tangent plane at the point (3, − 4, 5) Show that the tangent plane at any point on the cone in Exercise 1 passes through the EXERCISE 8 origin

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 how to plot z=9sqrt(x^2y^2) inside the Learn more about grpahSurf(x,y,z) I use this code and i can plot two graph 2 z functions but my teacher asked me to plot only the graph below the cone and i 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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 The top of the region (the orange colored surface) is the portion of the graph of the elliptic paraboloid\(z = 8 {x^2} {y^2}\) that is inside the cylinder \({x^2} {y^2} = 4\) The bottom of the region is the portion of the graph of the cone \(z = \sqrt {4{x^2} 4{y^2}} \) that is inside the cylinder \({x^2} {y^2} = 4\)Find the two parametric representation for the part of the sphere {eq}\displaystyle x^2 y^2 z^2 = 4 {/eq} that lies above the cone {eq}\displaystyle z = \sqrt{x^2 y^2 } {/eq} ParametricNow we save this in

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1) We have the surface {eq}z = \sqrt{x^2y^2} \iff z^2 = x^2 y^2\,(z\geq0) \iff x^2 y^2 z^2 =0\,(z\geq0) {/eq} This surface is a circular conePlot3D5 Sqrtx^2 y^2, {x, 5, 5}, {y, 5, 5}, RegionFunction > Function{x, y, z}, 0 < z < 5 An essential difference between RegionFunction and PlotRange when using RegionFunction, all points generated outside the region are discarded before building the 3D object to show, and the boundary of the region is computed and plotted nicely1269(a)Find and identify the traces of the quadric surface x2 y2 z2 = 1 and explain why the graph looks like the graph of the hyperboloid of one sheet in Table 1 x= k)k2 y2 z2 = 1 )y2 z2 = 1 k2 y= tsint, z= tlies on the cone z2 = x2 y2, and use this fact to help sketch the curve (tcost)2

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This problem has been solved!A) Find the curve of intersection of these for Teachers for Schools for Working Scholars9 Use spherical coordinates to find the volume of the solid that lies inside the sphere x^2y^2z^2=9, outside the cone z=sqrt (3x^23y^2) and above the xyplane 10 Evaluate triple integral (z) dV, where E region lying above the xyplane, under the graph of z=16x^2y^2, inside r=4sin (theta) and outside r=2sin (theta) Question 9

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Piece of cake Unlock StepbyStep Natural Language Math InputWe put this value along with the value we determined for #rho# into the conversion for #z# to find where the sphere and cone intersect #x^2y^2z^2=2# #(sqrt(x^2y^2))^2z^2=2# #z^2z^2=2# #2z^2=2# #z^2=1# #z=1=>z=1# The conversion for #z# is #z=rhocos(phi)# #=>rhocos(phi)=1# #cos(phi)=1/rho# As we know that #rho=sqrt2# #=>cos(phi)=1/sqrt(2)#Experts are tested by Chegg as specialists in their subject area We review their content and use your feedback to keep the quality high

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The cone z = sqrt (x^2 y^2) can be drawn as follows In cylindrical coordinates, the equation of the top half of the cone becomes z = r We draw this from r = 0 to 1, since we will later look at this cone with a sphere of radius 1 > cylinderplot ( r,theta,r,r=01,theta=02*Pi);Y 2 z = 0 Solution Traces in x= kare y2 z2 = 9k2, family of hyperbolas for k6= 0 and intersecting lines for k= 0 Traces in y= kare 9x2 z2 = k2;K 0, family of ellpises, traces in z= kare y2 9z2 = k2 again ellipses for k6= 0 Graph is an elliptic cone with axis the yaxis, vertex origin 2) Consider the equation y 2= x 4z2 4 Reduce toPlot sqrt(1 x y), sqrt(x^2 y^2 2 x y) Natural Language;

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Answer to The cone z = \\sqrt{x^{2} y^{2}} and the plane z = 1 y By signing up, you'll get thousands of stepbystep solutions to your homeworkThe lower bound z = x 2 y 2 z = x 2 y 2 is the upper half of a cone and the upper bound z = 18 − x 2 − y 2 z = 18 − x 2 − y 2 is the upper half of a sphere Therefore, we have 0 ≤ ρ ≤ 18, 0 ≤ ρ ≤ 18, which is 0 ≤ ρ ≤ 3 2 0 ≤ ρ ≤ 3 2 For the ranges of φ, φ, we need to find where the cone So, given a point in spherical coordinates the cylindrical coordinates of the point will be, r = ρsinφ θ = θ z = ρcosφ r = ρ sin ⁡ φ θ = θ z = ρ cos ⁡ φ Note as well from the Pythagorean theorem we also get, ρ2 = r2 z2 ρ 2 = r 2 z 2 Next, let's

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Easy as pi (e) Unlock StepbyStep Natural Language Math Input NEWUse textbook math notationGiven The Cone S 1 Z Sqrt X 2 Y 2 And The Hemisphere S 2 Z Sqrt 2 X 2 Y 2 A Find The Curve Of Intersection Of These Surfaces B Using Cylindrical For more information and source, see onExample Find the volume of the solid region above the cone z2 = 3(x2 y2) (z ≥ 0) and below the sphere x 2 y 2 z 2 = 4 Soln The sphere x 2 y 2 z 2 = 4 in spherical coordinates is ρ = 2

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Find the surface area of the portion of the cone z=sqrt(x^2y^2) in between the planes z=2 and z=3 I believe the answer should be 3sqrt2*pi Please show detailed work Question Find the surface area of the portion of the cone z=sqrt(x^2y^2) in between the planes z=2 and z=3 I believe the answer should be 3sqrt2*piBecause the's square root of for X squared plus y squared Okay, so the next thing we're gonna do is just have our graph ready This is the X This is the why this is the you see So as we conceive, our domain is not really defined at anyAll right So for this problem, doing the same thing we're discussing the graph about function in this gonna be And except why?

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Z=x^2y^2 WolframAlpha Volume of a cylinder?Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance The intersection of two surfaces will be a curve, and we can find the vector equation of that curve When two threedimensional surfaces intersect

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 Find the volume between the cone $y = \sqrt {x^2 z^2} $ and the sphere $x^2 y^2 z^2 = 49$ I know that the volume we're interested in is the volume of theZ=sqrt (x^2y^2) WolframAlpha Area of a circle?Plane z = 1 The trace in the z = 1 plane is the ellipse x2 y2 8 = 1

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See the answer Find the equation of the cone z=sqrt 3x^23y^2 in spherical polar coordinates Use spherical polar coordinates to evaluate the volume of the ice cream cone shaped region bounded below the cone z=sqrt 3x^23y^2 and above by the sphere x^2y^2z^2

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