(a) Sgiven by z= x2y2, 1 x 1, 1 y 1 oriented positive up Calculate RR S Fd~ S~ for F~= x~i y~j z~k (b) Sis the surface of 4x2 4y2 z2 6z 5 = 0 oriented inward Calculate the surface area of S (c) S is the surface of intersection of the sphere x2 y2 z2 4 and the plane z = 1 oriented away from the origin Calculate theFind 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=1The sphere x2 y2 z2 = 4 and the coordinate planes using Spherical Coordinates G xy⋅⋅z V ⌠ ⎮ ⌡ d Page 13 of 18
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The part of the sphere x^2+y^2+z^2=4-1 Find a parametrization of the portion of the sphere x2 y2 z2 = 8 in the rst octant between the xyplane and the cone z= p x 2 y like x166 #6 Sol Consider in cylindrical coordinate, then the sphere has parametric equation as r(r;Question Find the center and radius of a sphere, x^2y^2z^24x2y6z10=0 Found 2 solutions by Alan3354, Theo Answer by Alan3354() ( Show Source )
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Find the equation in spherical coordinates of $x^2 y^2 – z^2 = 4$ $$\begin{align} x^2 y^2 &= r^2\sin^2(\theta)\\ z^2 &= r^2 \cos(\theta) \\ x^2 y^2 z^2&=r^2(\sin^2(\theta) \cos^2(\theta)) = 4 \end{align}$$ Thus, $$r^2(\cos(2\theta)) = 4$$Math Advanced Math Q&A Library Find the extreme values of the function f (x, y, z) = xy z^2 on the circle in which the plane y x =0 intersects the sphere x^2 y^2 z^2 = 4Subject to the constraint g(x, y, z) = 4, where g(x, y, z) = x 2 y 2 z 2 Using Lagrange multipliers the equations for critical points We rewrite the first three equations Now we multiply the constraint equation by (1 − λ) 2 to obtain Note that we only need the value of 1
Find stepbystep Calculus solutions and your answer to the following textbook question Let S be the part of the sphere $$ x^2y^2z^2=25 $$ that lies above the plane z=4 If S has constant density k, find the center of mass about the zaxis6 The centre of the sphere whose equation X 2 y2 z2 2 X 4 Y 6 z 16 0 1S a 1 I I 6 the centre of the sphere whose equation x 2 y2 z2 2 School Maadi STEM School for Girls;The sphere x2 y2 z2 = 4 is the same as ˆ= 2 The cone z = p 3(x2 y2) can be written as ˚= ˇ 6 (2) So, the volume is Z 2ˇ 0 Z ˇ=6 0 Z 2 0 1 ˆ2 sin˚dˆd˚d 5 Write an iterated integral which gives the volume of the solid enclosed by z2 = x2 y2, z= 1, and z= 2 (You need not evaluate) x
Please Subscribe here, thank you!!!The plane x 2y – z = 4 cuts the sphere x2 y2 z2 – x z – 2 = 0 in a circle of radius (a) 1 (b) 3 (c) 2 (d) 2 Welcome to Sarthaks eConnect A unique platform where students can interact with teachers/experts/students to get solutions to their queriesEquation 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 ∫x (r,θ,𝛾) = r cosθ ∫y (r,θ,𝛾) = r sinθ ∫z (r,θ,𝛾) = 𝛾
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2x^2 2y^2 2z^2 2x 4y 2z 3 = 0 Solve Study Textbooks Guides Join / Login Question Find the centre and the radius of the sphereFind the equation of the sphere which passes through the circle x^2 y^2 = 4 , z = 0 and is cut by the plane x 2y 2z = 0 in a circle of radius 3Problem 23 Medium Difficulty Find a parametric representation for the surface The part of the sphere $ x^2 y^2 z^2 = 4 $ that lies above the cone $ z = \sqrt{x^2 y^2} $
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Let S1 be the sphere x 2 (y 1) 2 (z − 1) 2 =4, and let S2 be the sphere x 2 y 2 − 6y z 2 − 2z= 6 (a) Find the center and radius of each sphere (b) As carefully as you can, graph both spheres on a single set ofxyzaxes (c) The intersection of the two spheres is a circle, and thatcircle sits in a vertical plane y = y0 Find y0Equation of sphere x^2y^2z^2 = constant , volume element in spherical polars is dV = r^2 sin (theta) dr d (theta) d (phi) (a) i sketched a quarter of a sphere centred at x=0 , y=2 , z=0 (b ) ∫ ∫ √ (4x 2 (y2) 2) dx dy with limits 0 <Math Calculus University Calculus Early Transcendentals (4th Edition) The point on the sphere x 2 ( y − 3 ) 2 ( z 5 ) 2 = 4 nearest the xy plane The point on the sphere x 2 ( y − 3 ) 2 ( z 5 ) 2 = 4 nearest the xy plane Question Chapter 111, Problem 71E (a) To determine
If The Maximum And Minimum Distances Of The Sphere X 2 Y 2 Z 2 4x 8y 6z 4 0 From The Point 0 10 0 Are Alpha Alpha1 Respectively Then The Value Of Alpha 2 22alpha1 Equals
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Course Title MATH 143;ExampleFind the centroid of the solid region E lying inside the sphere x2y2z2 = 2z and outside the sphere x2 y2 z2 = 1 Soln By the symmetry principle, the centroid lies on the z axis Thus we only need to compute ¯z The top surface is x2 y2 z2 = 2z ⇒ ρ2 = 2ρcos(φ) or ρ = 2cos(φ) The bottom surface is x2 y2 z2 = 1 ⇒ ρThe Sphere x^2 y^2 z^2 = 4 intersects the plane z= x2 along a circle C in R3 Let the sphere be oriented clock wise as observed from above the zaxis Determine the curve integral Show transcribed image text Expert Answer Transcribed image text ST F dr С y2 F = (xy zy, xy) 9 2 Previous question Next question
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Find the area of the part of the sphere $ x^2 y^2 z^2 = 4z $ that lies inside the paraboloid $ z = x^2 y^2 $ Answer $4 \pi$ View Answer More Answers 0011 Frank L Related Courses Calculus 3 Calculus Chapter 16 Vector Calculus Section 6//googl/JQ8NysFind a Unit Normal Vector to the Sphere x^2 y^2 z^2 = 57 at (4, 4, 5)#x^2 y^2 z^2 = 4# The point given has the same magnitude in each dimension 1 = 1 = 1 Since they must be colinear, the desired farthest point must also have the same magnitude in each dimension In other words;
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Click here👆to get an answer to your question ️ Find the centre and the radius of the sphere x^2 y^2 z^2 2x 4y 6z 5 = 0 , &General Equation of sphere covering the given circle is x^2y^2z^22x3y4z6k(3x4y5z15)=0 We need a proper condition to solve for k It may cut the third sphere (a) touch it or (b) it may pass through center of second sphere or (c) it may pass through any two points of the second sphere or whateverThe cylinder is given by x 2 y 2 = 1 That will intersect the sphere when x 2 y 2 z 2 = 1 z 2 = 2 that is, when z= 1 and 1 so the cylinder has height 2 The volume of a sphere with radius is The volume of a cylinder with radius 1 and height 2 is
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I Limits in y 0 6 y 6 √ 4 − x2, so the positive side of the disk x2 y2 6 4 I Limits in z 0 6 z 6 p 4 − x2 − y2, so a positive quarter of the ball x2 y2 z2 6 4 2 z x y 2 2 Triple integral in spherical coordinates Example Change to spherical coordinates and compute the integral I = Z 2 −2 Z √ 4−x2 0 Z √ 4−x2−y2 0 y pSolution to Problem Set #9 1 Find the area of the following surface (a) (15 pts) The part of the paraboloid z = 9 ¡In this activity we work with triple integrals in cylindrical coordinates Let S be the solid bounded above by the graph of z = x 2 y 2 and below by z = 0 on the unit disk in the x y plane The projection of the solid S onto the x y plane is a disk Describe this disk using polar coordinates
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Find stepbystep Calculus solutions and your answer to the following textbook question ∫∫s y^2dS S is the part of the sphere $$ x^2y^2z^2=4 $$ that lies inside the cylinder $$ x^2y^2=1 $$ and above the xyplaneLet (3,4,−1) and (−1,2,3) be the end points of a diameter of a sphere Then, the radius of the sphere is equal to MediumOn the sphere x^2y^2z^2 = 13^2, there are many great circles that intersect at (3, 4, 12) At what points does the helix r = sin(t), cos(t), t intersect the sphere x 2 y 2 z 2 = 37?
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A sphere is inscribed in the tetrahedron whose faces are x = 0, y = 0, z = 0 and 2 x 6 y 3 z − 1 4 = 0 Find its centre and radius This question has multiple correct optionsThe equation of the sphere concentric with the sphere x 2y 2z 2−4x−2y−6z− 7=0 and passing through (0,0,0) is Medium View solution >Uploaded By BaronWombat966 Pages 25 This preview shows page 18 21 out of 25 pages
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2 and 0 <How to find the center, radius, and equation of the sphere The formula for the equation of a sphere We can calculate the equation of a sphere using the formula ( x − h) 2 ( y − k) 2 ( z − l) 2 = r 2 (xh)^2 (yk)^2 (zl)^2=r^2 ( x − h) 2 ( y − k) 2 ( z − l) 2 = r 2 where ( h, k, l) (h,k,l) ( h, k, l) is the center ofFind stepbystep Trigonometry solutions and your answer to the following textbook question find the center and radius of the sphere $$ x^2y^2z^24x8z19=0 $$
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`x^2 (y2)^2 (z3)^2 = (sqrt 17)^2` This equation looks like the standard form of the equation of the sphere with center (0 ;(Round your answers to three decimal places Enter your answers from smallest to largest zvalue)4 Evaluate RRR H e p x 2y z dV, where His enclosed by sphere x2 y2 z2 = 9 in the rst octant Solution In spherical coordinates it becomes R ˇ=2 0 R ˇ=2 0 R 3 eˆˆ2 sin˚dˆd˚d = (ˇ=2)(5e3 2) 5 Find the volume of the solid that lies within the sphere x 2 y2 z = 4, above the xyplane and below the cone z= p x2 y2
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#x=y=z# We can therefore rewrite the function for our sphere as;Y2 that lies above the x¡y plane ±4 ±2 0 2 4 x ±4 ±2 0 2 4 y ±4 ±2 0 2 4 Solution The part of the paraboloid z = 9¡x2 ¡y2 that lies above the x¡y plane must satisfy z = 9¡x2 ¡y2 ‚ 0 Thus x2 y2 • 9 We3) and the radius `sqrt 17` Actually, the redius of the
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Answer (1 of 3) x^2y^2=1 is a cylinder x^2 y^2 z^2 =4 is a sphere → z=(4(x^2 y^2))^(1/2) V=∫∫∫dxdydz V=∫∫ dxdy ∫dz from z=(4(x^2 y^2))^(1/2) toX2 y2 Solution x = rcosθ, y = rsinθ, z = z, dV = rdrdθdz ZZZ E z dV = Z2π 0 Z1 0 Zr r2 zrdzdrdθ = Z2π 0 Z1 0 z2r 2 z=r z=r2 drdθ = Z2π 0 Z1 0 r3 2 − r5 2 drdθ = Z2π 0 r4 8 − r6 12 r=1 r=0 dθ = Z2π 0 1 24 dθ = π 12 Problem 5 Evaluate RRR E y dV, where E is enclosed by the sphere x2 y2 z2 = 1 in the first octantEquation of the sphere with center in the positive octant which passes through the circle x 2 y 2 = 4, z = 0 and is cut by the plane x 2 y 2 z = 0 in a circle of radius 3 is Medium View solution
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#x^2 x^2 x^2 = 4# And solve for x #3x^2 = 4# #x^2 = 4/3#41,847 969 It's not really necessary to use calculus at all!) = (rcos )i (rsin )j p 8 r2 k Substituting z = p x 2 y 2into x y z2 = 8, we get x 2 y = r = 4
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B = 1 −2cz = 4z c = − 2 The standard equation of the first sphere is (x −( − 2))2 (y −1)2 (z − ( − 2))2 = 22 The center is ( − 2,1, − 2) and the radius, r = 2 The standard equation of second sphere is (x −0)2 (y −0)2 (z − 0)2 = 22 The center is (0,0,0) and the radius, R = 2Example Evaluate the surface integral IS = ∫ ∫ y2 dS where S is the part of the sphere x2 y2 z2 = 4 that lies inside the cylinder x2 y2 = 1 and above the xyplane See the figure below Strategy Use a two step process Step 1 Express f (x,y,z) in terms ofSince the surface is in the form x = f ( y, z) x = f ( y, z) we can quickly write down a set of parametric equations as follows, x = 5 y 2 2 z 2 − 10 y = y z = z x = 5 y 2 2 z 2 − 10 y = y z = z The last two equations are just there to acknowledge that we can choose y y and z z to be anything we want them to be
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