Follow 34 views (last 30 days) Show older comments Carter Pennington on Vote 0 ⋮ Vote 0 Commented Carter Pennington on clear all;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, music If we let `z=2` in `z=sqrt(x^2y^2)`, we eventually arrive at `x^2y^2=4`, which is a circle of radius 2, which we sketch at a height of `z=2` Putting together the traces and the parallel cross sections leads us to believe that the surface defined by `z=sqrt(x^2y^2)` is a cone
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Z=sqrt(4-x^2-y^2)
Z=sqrt(4-x^2-y^2)-Z = 6 is in the range of e^ sqrt 4 x^2 y^2 z = ln sqrt x^2 y^2 is a solution of LaPlace's equation If f(x, y, z) = e^ xyz^2 then fxy = xyz^4 e^ xyz^2 If z = 2x^2 2y^2, then an equation of the tangent plane at (1, 1, 4) is given by z 4x 4y 4 = 0I thought first that it might be a cone!
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The 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 zFigure \(\PageIndex{2}\) The graph of \(z=\sqrt{16−x^2−y^2}\) has a maximum value when \((x,y)=(0,0)\) It attains its minimum value at the boundary of its domain, which is the circle \(x^2y^2=16\) In Calculus 1, we showed that extrema of functions of one variable occur at critical points The same is true for functions of more than one 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 This is an elliptic paraboloid and is an example of a quadric surface We saw several of these in the previous section
Video Transcript All right So we need for X squared to the, uh, sorry Needs to be bigger than our ankle to zero We need nine minus y square to be bigger than or equal to zeroY=sqrt(4x^(2)z^(2)) y=sqrt(4x^(2)z^(2)) Expresiones con funciones; The value of f(3, ) is 1 √3 The domain of the function is y < ±2 while the range of the function exists in the range 1≤ f(x)≤3 Domain and range of a function Given the function expressed as To determine the value of f(3, 1) Hence the value of f(3, ) is 1 √3 THe domain is the point where the function exists
Tìm giới hạn \(\mathop {\lim }\limits_{(x,y) \to (0,0)} \frac{{{x^3}y}}{{{x^4} {y^4}}}\) Khảo sát cực trị của \(z = 1 \sqrt {{{(x 1)}^2} {y^2 $\tiny{}$ Find the volume of the given solid region bounded below by the cone $z=\sqrt{x^2y^2}$ and bounded above by the sphere $x^2y^2z^2=128$Find stepbystep Calculus solutions and your answer to the following textbook question Find the area of the surface The part of the sphere x^2y^2z^2=4 that lies above the plane z=1
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Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more 4 0 Here is the problem First Part (already done) Find the volume of the solid that is bounded above by the cylinder , on the sides by the cylinder , and below by the xyplane Answer Using the integral worked out above, and assuming that Setup the integral to find the average value of the function within that solidF(x, y, z) = (xz^2, x^2 y z^3, 2xy y^2 z), M is the outwardoriented boundary of the solid bounded by z = \sqrt {a^2 x^2 y^2} and z = 0 View Answer Let
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Y es igual a raíz cuadrada de (4 menos x en el grado (2) menos z en el grado (2)) y es igual a raíz cuadrada de (cuatro menos x en el grado (dos) menos z en el grado (dos)) y=√(4x^(2)z^(2)) y=sqrt(4x(2)z(2)) y=sqrt4x2z2;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 the curve of intersectionIn other words, a number y whose square (the result of multiplying the number by itself, or y ⋅ y) is x For example, 4 and −4 are square roots of 16, because 4 2 = (−4) 2 = 16 Every nonnegative real number x has a unique nonnegative square root, called the principal square root, which is denoted by , where
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Set up, but do not evaluate, an iterated integral expression in spherical coordinates whose value is the mass of the solid obtained by removing the cone \(\phi=\frac{\pi}{4}\) from the sphere \(\rho = 2\) if the density \(\delta\) at the point \((x,y,z)\) is \(\delta(x,y,z) = \sqrt{x^2y^2z^2}\text{}\)Graph z= sqrt(4y(x^2y^2)) sqrt(x^2y^2) Natural Language; Section 45 Triple Integrals Now that we know how to integrate over a twodimensional region we need to move on to integrating over a threedimensional region
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what is the simplified form of 3 sqrt (5c) x sqrt (15c^3) choices are a 15c^2 sqrt (3) b6c^2 sqrt (5) c 5c^2 sqrt(3) d 12c^4 sqrt(5) please help i don't understand how to do the question i would really appreciate itGiải hệ phương trình a) \(\hept{\begin{cases}x^2y^2=1\\x^9y^9=1\end{cases}}\) b) \(\hept{\begin{cases}\sqrt{x}\sqrt{y}\sqrt{z}=14\\\frac{1}{3x2y how to plot z=9sqrt(x^2y^2) inside the cylinder r=2?
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Example 1541 We find the area of the hemisphere z = 1 − x 2 − y 2 We compute the derivatives f x = − x 1 − x 2 − y 2 f y = − y 1 − x 2 − y 2, and then the area is ∫ − 1 1 ∫ − 1 − x 2 1 − x 2 x 2 1 − x 2 − y 2 y 2 1 − x 2 − y 2 1 d y d x This is a bit on the messy side, but we can use polarExample 1671 Suppose a thin object occupies the upper hemisphere of $x^2y^2z^2=1$ and has density $\sigma(x,y,z)=z$ Find the mass and center of mass of the z=sqrt(2x^y^2) that looks like a paraboloid?!
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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 dr Select a Web Site Choose a web site to get translated content where available and see local events and offers Based on your location, we recommend that you selectSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
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Calculator Use Enter 2 sets of coordinates in the 3 dimensional Cartesian coordinate system, (X 1, Y 1, Z 1) and (X 2, Y 2, Z 2 ), to get the distance formula calculation for the 2 points and calculate distance between the 2 points Accepts positive or negative integers and decimals Note that solving for #y# gives #y=pmsqrt(4x^2)#, which is a set of two functions, since a circle by itself does not pass the vertical line test, so a circle is not a function but can be described by a set of #2# functions Thus #y=sqrt(4x^2)# is the top half of the circle, which starts at #(2,0)#, rises to #(0,2)#, then descends to #(2,0)#, showing its range of #0lt=ylt=2# Functions Item Description POW Function Computes the value of the first argument raised to the value of the second argument SQRT Function Computes the square root of the input parameter Input value can be a Decimal or Integer literal or a reference to a column containing numeric values All generated values are nonnegative
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2 We can solve the integral ∫ x 2 4 d x \int\sqrt {x^24}dx ∫ x2 4 dx by applying integration method of trigonometric substitution using the substitution x = 2 tan ( θ) x=2\tan\left (\theta \right) x = 2tan(θ) Intermediate steps As Danny says, you should write the equation of the sphere as \(\displaystyle z = \sqrt{4x^2y^2}\) To find the region over which to integrate, you need to know where the sphere meets the cone, namely where \(\displaystyle \sqrt{x^2y^2} = \sqrt{4x^2y^2}\) or in other words \(\displaystyle x^2y^2=2\)1−x2−z2/4 0 f(x,y,z)dydxdz Solution Since y ranges from 0 to y = 3 p 1− x2 −z4/4, we have the upper surface y2 9 x2 z2 4 = 1, which is an ellipsoid We also note that the projected region R in the x−z plane has goes between x = 0 and x = p 1− z2/4, the latter being the boundary of an ellipse, while z ranges from 0 to 2 Therefore,
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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 $2$ Your inferior limit for $z$ is the lower cone and superior limit is the upper cone EDIT You seem to be getting the bounded volume by the cones wrong It is not what you drew with dashesEquation of cylinder is x 2 y 2 = 1 Equation of sphere is x 2 y 2 z 2 = 4 In cartesian coordinates, the region is given by 1 ≤ x ≤ 1, √1 x 2 ≤ y ≤ √1 x 2 and √4 x 2 y 2 ≤ y 2 ≤ √4 x 2 y 2 Converting to cylindrical coordinates, using We get a Jacobian determinant of r, 2 ≤ z ≤ √4 r 2 UsingSolution First, we will calculate fx(x, y) and fy(x, y), then we'll calculate the required tangent plane equation using the general equation z = f(xo, yo) fx(xo, yo)(x − xo) fy(xo, yo)(y − yo) with xo = π 3 and yo = π 4 fx(x, y) = 2cos(2x)cos(3y) fy(x, y) = − 3sin(2x)sin(3y) f(π 3, π 4) = sin(2(π 3))cos(3(π 4)) = (√3 2
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Answer (1 of 7) \iint \frac{x}{\sqrt{x^2y^2}}\ dxdy =\int\left( \int \frac{x}{\sqrt{x^2y^2}}\ dx\right)\ dy =\int\left( \frac12\int \frac{2x}{\sqrt{x^2y^2}}\ dxIntegral of sqrt(4 x^2) dxWatch more videos at https//wwwtutorialspointcom/videotutorials/indexhtmLecture By Er Ridhi Arora,Experts are tested by Chegg as specialists in their subject area We review their content and use your feedback to keep the quality high 100% (1 rating)
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In mathematics, a square root of a number x is a number y such that y 2 = x;Detailed step by step solution for ln((\sqrt3{x}sqrt(y^2))/(z^4)) This website uses cookies to ensure you get the best experience By using this website, you agree to our Cookie PolicyFind stepbystep Calculus solutions and your answer to the following textbook question Find the area of the surface The portion of the cone z = 2√x² y² inside the cylinder x² y² = 4
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Free PreAlgebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators stepbystepExplain why z=sqrt(4x^2y^2) is a graph of function, but x^2y^2z^2=4 is not Expert Answer Who are the experts? Proof Recall that the ring of Gaussian integers is a Euclidean Domain with respect to the norm N ( a b i) = a 2 b 2 for a b i ∈ Z i In particular, Ring of Gaussian Integers and Determine its Unit Elements Denote by i the square root of − 1 Let R = Z i = { a i b ∣ a, b ∈ Z }
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Reply Answers and Replies #2 caz Gold Member 708 755 Try rearranging the equation to (2x)=y 2 z 2 and look at it in the plane z=0 or y=0 Last edited Reply Likes AmaelleAnswer (1 of 4) Cones, just like spheres, can be easily defined in spherical coordinates The conversion from cartesian to to spherical coordinates is given below x=\rho sin\phi cos\theta y=\rho sin\phi sin\theta z=\rho cos\phi, where \phi is theVideo Transcript So for this problem, we're going to be graphing A multi variable function on Day One we're given is thes square root of four minus four X squared Mhm That's going to be minus y squared So a graph this here on, we see that this is what we end up getting is our graph We could try to extend these say, negative three and
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Error is the surface described should be z=sqrt (4x^2y^2) Image transcription text 3 Consider the "icecream cone" region between the graphs of z V (1 / 3X32 yz) and z V4 — m2 We want to find the center of mass of this region, with density function 6 (3, 3;, z) = z (a) (1 Point) Explain (in a few sentences) why 5 =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
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