I am already using it and I only can plot in 2 dimensional graph Can someone help me with this problem?Solution The graph of f in R3 can be realized as the level surface (or hypersurface) S = {(x,y,z) ∈ R3 g(x,y,z) = 0}, g(x,y,z) = x2 y2 z −9 Our point is then (1,−2,4) on the level surface The general formula for the tangent plane T pS to a point p onCircle on a Graph Let us put a circle of radius 5 on a graph Now let's work out exactly where all the points are We make a rightangled triangle And then use Pythagoras x 2 y 2 = 5 2 There are an infinite number of those points, here are some examples
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X^2 y^2 z^2=r^2 graph
X^2 y^2 z^2=r^2 graph-Answer (1 of 3) It's the equation of sphere The general equation of sphere looks like (xx_0)^2(yy_0)^2(zz_0)^2=a^2 Where (x_0,y_0,z_0) is the centre of the circle and a is the radious of the circle It's graph looks like Credits This 3D Graph is created @ code graphing calculatorPlotting graphics3d Share Improve this question Follow asked Nov 29 '15 at 533 user user



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For all x;y 2C, x 6= y, and 2(0;1), then we say that f(x) is strictly concave Intuitively, the graph of a convex function lies on or below any chord between two points on the graphSurface area of x^2 y^2 z^2 = r^2 So I tried to integrate the function for a sphere to find the surface area and rotate it to find the volume, not hypervolume, of a hypersphere I tried a few times and kept getting (128/3)×𝜋×r 3 ×i×r 3 I used the equation for the surface area a 3D graph, S=∬ (1 (dz/dx) 2 (dz/dy) 2 dX dy I do not know what I did wrong or if my equations are wrongGraph y^2z^2=9 y2 z2 = 9 y 2 z 2 = 9 Reorder terms x2 y2 = 9 x 2 y 2 = 9 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard form The variable r r represents the radius of the
P ∈ R3 is the ordered triple (r,θ,z) defined by the picture y z x 0 P r z Remark Cylindrical coordinates are just polar coordinates on the plane z = 0 together with the vertical coordinate z Theorem (Cartesiancylindrical transformations) The Cartesian coordinates of a point P = (r,θ,z) are given by x = r cos(θ), y = r sin(θ), and z = zGraph x^2y^2=1 x2 − y2 = −1 x 2 y 2 = 1 Find the standard form of the hyperbola Tap for more steps Flip the sign on each term of the equation so the term on the right side is positive − x 2 y 2 = 1 x 2 y 2 = 1 Simplify each term in the equation in order to set the right side equal to 1 1 The standard form of anFirst notice the graph of the surface z = 27 − 2 x 2 − y 2 z = 27 − 2 x 2 − y 2 in Figure 59(a) and above the square region R 1 = −3, 3 × −3, 3 R 1 = −3, 3 × −3, 3 However, we need the volume of the solid bounded by the elliptic paraboloid 2 x 2 y 2 z = 27, 2 x 2 y 2 z = 27, the planes x = 3 x = 3 and y = 3
A sphere is the graph of an equation of the form x 2 y 2 z 2 = p 2 for some real number p The radius of the sphere is p (see the figure below) Ellipsoids are the graphs of equations of the form ax 2 by 2 c z 2 = p 2 , where a , b , and c are all positiveWolframAlpha Computational Intelligence Natural Language Math Input NEW Use textbook math notation to enter your math Try it × Extended Keyboard Examples Compute expertlevel answers using Wolfram's breakthrough algorithms, knowledgebase and AI technologyDe ned by z= x 2 y2 and x 2y2 z2 = 7 at the point ( 1;1;2) Hint Think about the geometry of the gradient vectors You don't have to parametrize the curve to do this problem Solution The surface z= x2 y2 can be written as the level surface F(x;y;z) = x 2 y z= 0;



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Conic Sections (see also Conic Sections) Point x ^2 y ^2 = 0 Circle x ^2 y ^2 = r ^2 Ellipse x ^2 / a ^2 y ^2 / b ^2 = 1 Ellipse x ^2 / b ^2 y ^2 / a ^2 = 1 Hyperbola x ^2 / a ^2 y ^2 / b ^2 = 1 Parabola 4px = y ^2 Parabola 4py = x ^2 Hyperbola y ^2 / a ^2 x ^2 / b ^2 = 1 For any of the above with a center at (j, k) instead of (0,0), replace each x term with (xj) andPlot z=x^2y^2 WolframAlpha Have a question about using WolframAlpha?If one of the variables x, y or z is missing from the equation of a surface, then the surface is a cylinder Note When you are dealing with surfaces, it is important to recognize that an equation like x2 y2 = 1 represents a cylinder and not a circle The trace of the cylinder x 2 y = 1 in the xyplane is the circle with equations x2 y2



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Where Ris the region that lies to the left of the yaxis between the circles x2 y2 = 1 and x2 y2 = 4 Solution This region Rcan be described in polar coordinates as the set of all pointsGraph y= (x2)^2 y = (x − 2)2 y = ( x 2) 2 Find the properties of the given parabola Tap for more steps Use the vertex form, y = a ( x − h) 2 k y = a ( x h) 2 k, to determine the values of a a, h h, and k k a = 1 a = 1 h = 2 h = 2 k = 0 k = 0 Since theGraphs (z= f(x;y)) The graph of f R2!R is f(x;y;z) 2R3 jz= f(x;y)g Example When we say \the surface z= x2 y2," we really mean \The graph of the function f(x;y) = x2 y2" That is, we mean the set f(x;y;z) 2R3 jz= x2 y2g Level Sets (F(x;y;z) = c) The level set of F R3!R at height cis f(x;y;z) 2R3 jF(x;y;z) = cg



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To minimize f(x;y;z) = x 2 y2 z (the reason this is equivalent is that the square root function 1 2 JAMES MCIVOR is an increasing function, but you don't need to mention this, since we used this trick in lecture a few times already) To apply the extreme value theorem we need to consider a closed and bounded θ y = r sin θ r 2 = x 2 y 2 We are now ready to write down a formula for the double integral in terms of polar coordinates ∬ D f (x,y) dA= ∫ β α ∫ h2(θ) h1(θ) f (rcosθ,rsinθ) rdrdθ ∬ D f ( x, y) d A = ∫ α β ∫ h 1 ( θ) h 2 ( θ) f ( r cos All the surfaces have been the graph of some quadratic relation in $x,\, y,$ and $z$ like $z x^2 y^2 = 0$ in the case of a hyperbolic paraboloid or $x^2 y^2 z^2 = r^2$ for a sphere, all the crosssections of these surfaces have been conic sections like parabolas, hyperbolas etc In view of the first of these comments we make the following



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Algebra Graph x^2y^2=1 x2 y2 = 1 x 2 y 2 = 1 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard form The variable r r represents the radius of the circle, h h represents the xoffset from the origin, and k k represents the yoffset from originAnd so the gradient of Fis rF(x;y;z) = h2x;2y;Plane z = 1 The trace in the z = 1 plane is the ellipse x2 y2 8 = 1



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2 SURFACES Definition A subset S R3 is a regular surface if, for each point p S , there is an open neighborhood V of p in R3, an open set U R2 and a map X U V S , such that (1) X is smooth, meaning that if we write X(u, v) = (x(u, v), y(u, v), z(u, v)) , then the realvalued functions x(u, v) , y(u, v) and z(u, v)Trigonometry Graph x^2y^2=r^2 x2 y2 = r2 x 2 y 2 = r 2 Move all terms containing variables to the left side of the equation Tap for more steps Subtract r 2 r 2 from both sides of the equation x 2 y 2 − r 2 = 0 x 2 y 2 r 2 = 0 Move y 2 y 2 x 2 − r 2 y 2 = 0 x 2 r 2 y 2 = 0This tool graphs z = f (x,y) mathematical functions in 3D It is more of a tour than a tool All functions can be set different boundaries for x, y, and z, to maximize your viewing enjoyment This tool looks really great with a very high detail level, but you may find it more comfortable to use less detail if you want to spin the model



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r**2 == 1 and since r should be real, r == 1 (for any phi) Plugging this into python import numpy as np import matplotlibpyplot as plt phi = nplinspace(0, 2*nppi, 0) r = 1 x = r*npcos(phi) y = r*npsin(phi) pltplot(x,y) pltaxis("equal") pltshow()The cone is of radius 1 where it meets the paraboloid Since z = 2 − x 2 − y 2 = 2 − r 2 z = 2 − x 2 − y 2 = 2 − r 2 and z = x 2 y 2 = r z = x 2 y 2 = r (assuming r r is nonnegative), we have 2 − r 2 = r 2 − r 2 = r Solving, we have r 2 r − 2 = (r 2) (r − 1) = 0 r 2 r − 2 = (r 2) (r − 1) = 0 Since r ≥ 0, r ≥ 0, we have r = 1 r = 1 Therefore z = 1 z = 1Show activity on this post 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 Note that z = y 2 − x 2 is the semicone with z > 0, ie above the plane ( x, y) and z = − y 2 − x 2 is the semi



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Answer (1 of 3) The same way you plot anything Even with this equation being complicated looking, just assume that this elliptical mapping has some yvalue(s) for whatever xvalue(s) Since this is second order, we can expect it to have some values So, start off by making aSteps to graph x^2 y^2 = 4The paraboloid z = 9 − x2 − y2 below the xyplane and outside the cylinder x2 y2 = 1 Solution First sketch the integration region y x y =1 z z = 9 x y2 2 2 x 1 3 In cylindrical coordinates, z = 9 − x2 − y2 ⇔ z = 9 − r2 x2 y2 = 1 ⇔ r = 1 V = Z 2π 0 Z 3 1 Z 9−r2 0 r dz dr dθ = 2π Z 3 1 (9 − r2) r dr V = 2π 9r2



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Contact Pro Premium Expert Support » Give us your feedback »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, musicExtended 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, music



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2 We can describe a point, P, in three different ways Cartesian Cylindrical Spherical Cylindrical Coordinates x = r cosθ r = √x2 y2 y = r sinθ tan θ = y/x z = z z = z Spherical Coordinates x = ρsinφcosθ ρ = √x2 y2 z2 y = ρsinφsinθ tan θ = y/x z = ρcosφ cosφ = √x2 y2 z2 zMath Input NEW Use textbook math notation to enter your math Try itFor example, the graph of paraboloid 2 y = x 2 z 2 2 y = x 2 z 2 can be parameterized by r (x, y, z) be a function with a domain that contains S For now, assume the parameter domain D is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation



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Assignment 7 Solutions Math 9 { Fall 08 1 (Sec 154, exercise 8) Use polar coordinates to evaluate the double integral ZZ R (x y)dA;Substitute r 2 = x 2 y 2 r 2 = x 2 y 2 into equation r 2 z 2 = 9 r 2 z 2 = 9 to express the rectangular form of the equation x 2 y 2 z 2 = 9 x 2 y 2 z 2 = 9 This equation describes a sphere centered at the origin with radius 3 3 (see the following figure)3D plot x^2y^2z^2=4 Natural Language;



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Answer (1 of 8) Assuming you're only working with real numbers Rearange to get that x^2y^2=0^2 This is a circle of radius 0 cenetered the orgin But if our circle is of radius 0 and at the origin, that must mean one thing the graph is just the origin SoNext, let us draw the cylinder x^2 y^2 = 21;z 1) and Q(x 2;y 2;z 2) is jPQj= p (x 2 x 1)2 (y 2 y 1)2 (z 2 z 1)2 EXAMPLE 5 Show that the equation x2 y2 z2 2x 4y8z17 = 0 represents a sphere, and nd its center and radius In general, completing the squares in x2 y2 z2 Gx Hy Iz J= 0 produces an equation of the form (x a)2 (y b)2 (z c)2 = k If k>0 then the graph of



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X,y in RR rArr z^2=x^2y^2 Also, z^2=(xiy)^2=x^22ixyi^2y^2=x^2y^22ixy With these, the eqn becomes, x^2y^22x^22y^24ixy=2i, ie, 3y^2x^2i(4xy)=2i Explanation This is the equation of a circle with its centre at the origin Think of the axis as the sides of a triangle with the Hypotenuse being the line from the centre to the point on the circle By using Pythagoras you would end up with the equation given where the 4 is in fact r2 To obtain the plot points manipulate the equation as below Given x2 y2 = r2 → x2 y2 = 4In the twodimensional coordinate plane, the equation x 2 y 2 = 9 x 2 y 2 = 9 describes a circle centered at the origin with radius 3 3 In threedimensional space, this same equation represents a surface Imagine copies of a circle stacked on top of each other centered on the zaxis (Figure 275), forming a hollow tube



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For example, if we want to plot the top half of the sphere with equation x^2 y^2 z^2 = 16, we solve for z and obtain z = sqrt (16 x^2 y^2) or z = sqrt (16 r^2) Now we draw the graph parametrically, as follows > cylinderplot(r,theta,sqrt(16r^2),r=04,theta=02*Pi);(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;Where the two surfaces intersect z = x2 y2 = 8 − x2 − y2 So, 2x2 2y2 = 8 or x2 y2 = 4 = z, this is the curve at the intersection of the two surfaces Therefore, the boundary of projected region R in the x − y plane is given by the circle x2 y2 = 4 So R



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Solution Z 1 0 Z 2 2 2z Z x2z 2 0 f(x;y;z)dydxdz 4 (15points) Let Rbe the region in the xyplane that lies to the left of the yaxis and between the circles x2 y2 = 1 and x2 y2 = 4 Set up the integral ZZ R (xy)dA using polar coordinates Solution Z 3ˇ=2 ˇ=2 Z 2 1 (rcos rsin )rdrd = Z 3ˇ=2 ˇ=2 Z 2 1 r2(cos sin )drd 5 (15The graph of a 3variable equation which can be written in the form F(x,y,z) = 0 or sometimes z = f(x,y) (if you can solve for z) is a surface in 3D One technique for graphing them is to graph crosssections (intersections of the surface with wellchosen planes) and/or traces (intersections of the surface with the coordinate planes) How to plot 3 dimensional graph for x^2 y^2 = 1?



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