Thanks for contributing an answer to Mathematics Stack Exchange!Cones, just like spheres, can be easily defined in spherical coordinates The conversion from cartesian to to spherical coordinates is given below mathx=\rho sin\phi cos\theta/math mathy=\rho sin\phi sin\theta/math zmath=\rho cos\phi/mWe can calculate the following example problem Find the volume of cone of height 1 and radius one It is bounded by surface z = √x2 y2 and plane z = 1 The volume is ∫1 − 1∫√1 − x2 − √1 − x2∫1√x2 y2dzdydz The integral is easier to compute in cylindrical coordinates In cylindrical coordinates, the cone is described
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Graph z=sqrt(1-x^2-y^2)- Homework Statement Set up the volume integral of x^2 y^2 z^2 between z=sqrt(1x^2y^2) and z=x^2 Homework Equations Cartesian differential volume element, spherical differential volume element The Attempt at a Solution In class I've been told when doing triple integrals to tryGiven 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 Finding Volume Of Solid Under Z Sqrt 1 X 2 Y 2 Above The Region Bounded By X 2 Y 2 Y 0 Mathematics Stack Exchange
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history `z=sqrt(1(x^2y^2))` 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 true to find theZ=sqrt (x^2y^2) WolframAlpha Volume of a cylinder?
Z = √(1x 2 y 2), x = ln t, y=cost Expert Answer 100% (7 ratings) Previous question Next question Get more help from Chegg Solve it with our calculus problem solver and calculatorZ = sqrt(36 x^2 y^2) Let's rewrite that as the following z^2 = 36 x^2 y^2 x^2 y^2 z^2 = 36 This we know is a sphere, centered at the origin, of radius 6 However, sinc View the full answerFind the volume of the solid that is under the hemisphere $z=\\sqrt{1x^2y^2}$ above the region bounded by the graph of the circle $x^2 y^2y=0$ I solved this
Below the cone z = \sqrt{x^2 y^2} and above the ring 1 \le x^2 y^2 \le 4 Boost your resume with certification as an expert in up to 15 unique STEM subjects this summer Signup now to start earning your free certificateSurfaces and Contour Plots Part 2 Quadric Surfaces Quadric surfaces are the graphs of quadratic equations in three Cartesian variables in spaceThe problem now is
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Algebra Graph y = square root of 1x^2 y = √1 − x2 y = 1 x 2 Find the domain for y = √1 −x2 y = 1 x 2 so that a list of x x values can be picked to find a list of points, which will help graphing the radical Tap for more stepsPiece of cake Unlock StepbyStep Extended Keyboard ExamplesAnswer to Set up the integral in spherical coordinates \int_{0}^{1}\int_{0}^{\sqrt{1x^2}}\int_{\sqrt{x^2y^2}}^{\sqrt{1x^2y^2}} xydzdydx By
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history (delw)/(delx) = x/sqrt(x^2 y^2 z^2) (delw)/(dely) = y/sqrt(x^2 y^2 z^2) (delw)/(delz) = z/sqrt(x^2 y^2 z^2) Since you're dealing with a multivariable function, you must treat x, y, and z as independent variables and calculate the partial derivative of w, your dependent variable, with respect to x, y, and z When you differentiate with respect to x, you treat y and z asIt cannot be done Suppose to the contrary that it can be done We will derive a contradiction Suppose that \frac{x^2}{\sqrt{x^2y^2}}=f(x)g(y) for some functions f and g
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Answer to Given that z = sqrt(1 x^2 y^2), x = ln(t), and y = sin(t) Use the Chain Rule to find dz/dt (Give an answer in terms of t) ByThe Wigner semicircle distribution, named after the physicist Eugene Wigner, is the probability distribution on −R, R whose probability density function f is a scaled semicircle (ie, a semiellipse) centered at (0, 0) =for −R ≤ x ≤ R, and f(x) = 0 if x > R It is also a scaled beta distribution if Y is betadistributed with parameters α = β = 3/2, then X = 2RY – R has theFind the probability density function of the 'Magnitude" random variable in terms of the joint density function of its components For a generalization, see
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Answer to Evaluate the surface integral double integral over S of sqrt(1 x^2 y^2) dS where S is the surface z = sqrt(1 x^2 y^2) ByProblem 4 Find the volume of the solid bounded by z = 1−x2 −y2 and the xyplane A) 1 2 B) 2 3 C) 1 D) 4 3 ☎ E) π 2 F) 2π 3 G) π H) 4π 3 To find the volume we will integrate the height of the solid, over the projection of the solidContact Pro Premium Expert Support »
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Please be sure to answer the questionProvide details and share your research!Get stepbystep solutions from expert tutors as fast as 1530 minutes The region does not split up The shadow on the $xy$ plane is the circle where $z=4 \sqrt{x^2y^2}$ and $z=\sqrt{x^2y^2}$ meet This is when the radius is equal to
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A solid is bounded below by the cone {eq}z=\sqrt{x^2y^2}{/eq} and above by the plane {eq}z = 1{/eq} Find the center of mass and the moment of inertia about the {eq}zaxis{/eq} if the density isCompute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historySolution We use the formula m = ∬ S μ(x,y,z)dS The projection D(x,y) of the parabolic surface S onto the xy plane is the circle of radius 1 centered at the origin Hence, we can write m = ∬ S μ(x,y,z)dS = ∬ S zdS = ∬ D(x,y)(x2 y2)⋅√14x2 4y2dxdy By
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A quick video about graphing 3d for those who never done it before Pause the video and try itVideo Transcript {'transcript' "and this problem, we want to sketch the surface z equal to the square root of X squared plus y squared And so first, we want to know what kind of surface this is Eso if we look up at our forms So let's go ahead and square this and see what we get So we would have busy squared is equal to X squared plus y 1 I am trying to plot the following equation in MATLAB ratio = sqrt (11/ (kr)^2) With k and r on the x and y axes, and ratio on the z axis I used meshgrid to create a matrix with values for x and y varying from 1 to 10 x,y = meshgrid ( 1110, 1110);
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A lamina has the shape of a portion of sphere \(x^2 y^2 z^2 = a^2\) that lies within cone \(z = \sqrt{x^2 y^2}\) Let S be the spherical shell centered at the origin with radius a , and let C be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the z axisSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreFind the volume of the solid bounded by z= sqrt(x 2 y 2 )1 and x 2 y 2 z 2 = 25 Solve it by double integration method This question has been answered Subscribe to view answer Question Find the volume of the solid bounded by z= sqrt(x 2 2 2 2 2 y)1 and x y z = 25 Solve it by double integration method
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Multivariable Calculus Find the area of the surface z = (x^2 y^2)^1/2 over the unit disk in the xyplane After computing, we rederive the area formFinding Volume Of Solid Under Z Sqrt 1 X 2 Y 2 Above The Region Bounded By X 2 Y 2 Y 0 Mathematics Stack Exchange For more information and source, Given 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 In terms of our new function the surface is then given by the equation f (x,y,z) =0 f ( x, y, z) = 0 Now, recall that ∇f ∇ f will be orthogonal (or normal) to the surface given by f (x,y,z) =0 f ( x, y, z) = 0 This means that we have a normal vector to the surface The only potential problem is that it might not be a unit normal vector
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Question Let S be in the hemisphere z = sqrt 1x^2y^2 oriented with its normal vector n pointing upward Find the flux of the vector field F(x,y,z Show Solution Okay, since we are looking for the portion of the plane that lies in front of the y z y z plane we are going to need to write the equation of the surface in the form x = g ( y, z) x = g ( y, z) This is easy enough to do x = 1 − y − z x = 1 − y − z Next, we need to determine just what D D isGiven 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 on
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Let S be in the hemisphere z = sqrt 1x^2y^2 oriented with its normal vector n pointing upward Find the flux of the vector field F(x,y,z) =yi xj 3zk across the surface S That is calculate;X^2y^2z^2=1 WolframAlpha Have a question about using WolframAlpha?Take the square root of both sides of the equation x^ {2}y^ {2}z^ {2}=0 Subtract z^ {2} from both sides y^ {2}x^ {2}z^ {2}=0 Quadratic equations like this one, with an x^ {2} term but no x term, can still be solved using the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a}, once they are put in standard form ax^ {2}bxc=0
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Answer to Use the Chain Rule to find dz / dt (Enter your answer only in terms of t) z = \\sqrt{1 x^2 y^2}, x = ln(t), y = \\cos(t) dz / dt =But avoid Asking for help, clarification, or responding to other answers Example 1586 Setting up a Triple Integral in Spherical Coordinates Set up an integral for the volume of the region bounded by the cone z = √3(x2 y2) and the hemisphere z = √4 − x2 − y2 (see the figure below) Figure 15 A region bounded below by a cone and above by a hemisphere Solution
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