Find the volume of the region that lies under the paraboloid z = x 2 y 2 z = x 2 y 2 and above the triangle enclosed by the lines y = x, x = 0, y = x, x = 0, and x y = 2 x y = 2 in the x y x yplane (Figure 536) Solution First examine the region over which we need to set up the double integral and the accompanying paraboloidLet 𝒮 be the surface formed by the paraboloid z = 1x 2y 2, z ≥ 0, and the unit disk centered at the origin in the xy plane, graphed in Figure 1572, and let F → = 0, 0, z (This surface and vector field were used in Example 1563)Let Sbe the part of the paraboloid z= 7 x2 4y2 that lies above the plane z= 3, oriented with upward pointing normals Use Stokes' Theorem to nd ZZ S curlF~dS~ Solution Here is a picture of the surface S x y z The strategy is exactly the same as in#1 The
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Enclosed by the paraboloid z=x^2+y^2+1-We can also write the cone surface as r = z r = z and the paraboloid as r 2 = 2 − z r 2 = 2 − z The lower bound for r r is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid The plane z = 1 z = 1 divides the region into two paraboloide z=1x2y2 Publicado por tartari ( 2 intervenciones) el Gracias por contestarme, pues ya lo estaba haciendo así, pero no se porque razon la version de matlab que estoy utilizando no me muestra la grafica ni nada Además tambien debo realizar a continuacion las siguientes cosas Utilizando una parametrizacion
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Solution to Problem Set #9 1 Find the area of the following surface (a) (15 pts) The part of the paraboloid z = 9 ¡ x2 ¡ 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 WeAnswer to Find the area of the surface The part of the paraboloid z = 1 x^2 y^2 that lies above the plane z = 6 By signing up, you'll getLies beneath the paraboloid z = 1−x2 −y2 Solution The solid region E is 0 ≤ x ≤ 1, 0 ≤ y ≤ √ 1−x 2, 0 ≤ z ≤ 1−x2 −y So in cylindrical coordinates E 0 ≤ θ ≤ π2, 0 ≤ r ≤ 1, 0 ≤ z ≤ 1−r2 ZZZ E (x3 xy2)dV = Z π 2 0 Z 1 0 Z −r2 0 r3 cos3 θ (rcosθ)(r2 sin2 θ)rdzdrdθ = Z π 2 0 Z 1 0 Z −r2
Part Of The Elliptic Paraboloid Z X2 Y2 Which Can Be Generated By Stock Photo Alamy For more information and source, see on this link https Find The Area Of The Paraboloid Z 1 X 2 Y 2 That Lies In The First Octant Study Com For more information and source,Find the surface area of the part of the paraboloid z=16x^2y^2 that lies above the xy plane (see the figure below) The region R in the xyplane is the disk 05 (a) Find the center of mass of the solid S bounded by the paraboloid z = 4x2 4y2 and the plane z = 1 if S has constant density K Solution In cylindrical coordinates the region E is described by 0 ≤ r ≤ 1/2, 0 ≤ θ ≤ 2π, and 4r2 ≤ z ≤ 1 Thus, the mass of the solid is M = ZZZ E K dV = Z 2π 0 Z 1/2 0 Z 1 4r2 Krdzdrdθ = Kπ 8
Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history2 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 dr8 Find the surface area of the paraboloid z = 4 x2 y2 that lies above the xyplane Solution For this problem polar coordinates are useful S = ZZ D s 1 @z @x 2 @z @y 2 dA = ZZ D p 14x2 4y2 dA = Z2ˇ 0 Z2 0 r p 14r2 drd = Z2ˇ 0 1 12 (14r2)3=2 2 d = ˇ 6 (17)3=2 1 9 Find the surface area of the surface z = 2 3(x 3=2 y3=2) for 0 6 x
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Answer to Find the area of the paraboloid z = 1 x^2 y^2 that lies in the first octant By signing up, you'll get thousands of stepbystepFind The Volume Of The Solid Bounded By The Plane Z 0 And The Paraboloid Z 1 X 2 Y 2 Use A Double Integral And Polar Coordinates Study Com For more information and source, see on1 but if we instead describe the region using cylindrical coordinates, we nd that the solid is bounded
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyFind The Volume Of The Solid Below The Paraboloid Z 4 X 2 Y 2 And Above The Region R R Theta 0 Leq R Leq 1 0 Leq Theta Leq 2 Pi Study Com Find The Volume Of The Solid Bounded By The Paraboloid Z 1 X 2 Y 2 And Below The Plane Z 1 Y Hint Equate The Z Values Study Com For more information and source,Compute the mass of the 3D region under above the xyplane and below the paraboloid z = 1x 2y 2 which has a density function ρ (x, y, z) = ρ 0 x 2 y 2 (Hint Convert to Spherical or Cylindrical coordinates) 7 Find the surface area of the segment of the paraboloid x
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Sis the surface of the solid bounded by the paraboloid z= 1 x2 y2 and the xyplane ZZ S FdS= ZZZ E div FdV div F= 6x 2 3y2 3y = 6(x2 y2) ZZZ E 6(x2 y2) dV = 6 Z 2ˇ 0 Z 1 0 Z 1 r2 0 r2rdzdrd = 6 Z 2ˇ 0 Z 1 0 r3 z 1 2r 0 drd = 6Where Eis the solid bounded below by the paraboloid z= x2 y2, above by the plane z= 4, and the planes y= 0 and y= 2 This integral can be evaluated as an iterated integral Z 2 2 Zp 4 x2 0 Z 4 x 2y f(x;y;z)dzdydx; $\begingroup$ @saulspatz Well we want to find the SURFACE area of part of the paraboloid that lies above the plane z = 4 And there is a formula to calculating surface area as shown in my first picture $\endgroup$ – Not Friedrich gauss Apr 5 ' at 426
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The 2 given surfaces are reflections of each other at the plane y=z because each of them mapped onto the other by interchanging between y and z Therefore their intersection contained inside that plane, and it is the curve given by Hence the per 0 Evaluate the volume of V ⊂ R 3, which is bounded by paraboloid z = 1 − x 2 − y 2 and the surface z = 1 − y, for z ⩾ 0 Attempt The desired volume goes like ∬ D ( 1 − x 2 − y 2 − ( 1 − y)) d x d y, where D is the projection of V on R 2 How do we determine D?(pts) Let S be the surface formed by the part of the paraboloid z = 1−x2−y2 lying above the xyplane Orient S so that the normal vector is pointing upwards Let F~ = xˆıyˆ 2(1−z)kˆ (i) Find the flux of F~ across S directly Solution We have dS~ = h2x,2y,1idxdy
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Assignment 8 (MATH 215, Q1) 1 Use the divergence theorem to find RR S F ndS (a) F(x,y,z) = x3 i 2xz2 j 3y2z k;4 Find the volume and centroid of the solid Ethat lies above the cone z= p x2 y2 and below the sphere x 2y z2 = 1, using cylindrical or spherical coordinates, whichever seems more appropriate Recall that the centroid is the center of mass of the solidS is the surface of the solid bounded by the paraboloid z = 4 − x2 − y2 and the xyplane Solution The divergence of F is
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Solution The ux is the surface integral ZZ S F dS div F = 2xy2 3xy2 xy2 = 0 By the Divergence theorem ZZ S F d S = ZZZ E 0 dV = 0 2Solution If we put z = 0 in the equation of the paraboloid, we get x2 y2 = 1 This means that the plane intersects the paraboloid in the circle x2 y2 = 1, so the solid lies under the paraboloid and above the circular disk D given by x2 y2 ≤ 1 In polar coordinates D is given by 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π Since 1−x2 −y2 = 1−Find stepbystep solutions and your answer to the following textbook question Use cylindrical coordinates Evaluate triple integral (xyz)dV, where E is the solid in the first octant that lies under the paraboloid z=4x^2y^2
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Arrow_forward Buy Find launch Multivariable CalculusThe part of the paraboloid z = 1 x^2 y^2 that lies above the plane z = 2 Our Discord hit 10K members!Find the area of the surface The part of the paraboloid z = 4 − x 2 − y 2 z = 4 x^2 y^2 z = 4−x2 − y2 that lies above the xyplane Explanation A Explanation B Recall the formula for the area If
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The surface S of the solid Ebounded by the hyperboloid x2 y2 z2 = 1 and the paraboloid z= 4 2x2 y, when z 0 ) Have a wonderful day!Figure 1 Region S bounded above by paraboloid z = 8−x2−y2 and below by paraboloid z = x2y2 Surfaces intersect on the curve x2 y2 = 4 = z So boundary of the projected region R in the x−y plane is x2 y2 = 4 Where the two surfaces intersect z = x2 y2 = 8 − x2 − y2 So, 2x2 2y2 = 8 or x2 y2 = 4 = z, this is the curve atParaboloid z = x2 y2 and below the half cone z = p x2 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
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Use Stokes' Theorem to evaluate ∫ c F dr In each Case C is oriented counterclockwise as viewed from above 9 F(x, y, z) = xy i yz j zx k, C is the boundary of the part of the paraboloid z = 1 x 2 y 2 in the first octantProblem 3 Find the volume of the solid that lies between the paraboloid z = x2 y2 and the sphere x2 y2 z2 = 2 using 1 (15%) the cylindrical coordinate, and 2 (15%) the spherical coordinate Sol First we nd the intersection of the paraboloid and the sphere If (x,y,z) is on the intersection,It shows the paraboloid z = A x 2 B y 2 over the square domain1 ≤ x ≤ 11 ≤ y ≤ 1 If you change the domain to a disk, you will see the portion of the paraboloid for which 0 ≤ z ≤ 8 When you change A and B, the domain will change accordingly Here are a few things to think about
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Order my "Ultimate Formulmznto/2SKuojN Hire me for private lessons https//wyzantcom/tutors/jjthetutorRead "The 7 Habits of Successful STParaboloid z = x 24y A trigonometric parametrization will often be better if you have to calculate a surface integral ( u;v) = Here we want x2 4y2 to be simple So x = 2r cos y = r sin will do better Plug x and y into z = x2 4y2 to get the zcomponent10 Use cylindrical coordinates to evaluate ZZ E (x 3 xy 2) dV where E is the solid in the rst octant the lies beneath the paraboloid z = 1x 2y 2 11 Evaluate ZZZ E xe x 2 y 2 z 2 dV where E is the portion of the unit ball x 2 y 2 z 2 ≤ 1 that lies in the rst octant 7
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Problem 3 Let S be the boundary of the solid bounded by the paraboloid z = x2 y2 and the plane z = 4, with outward orientation (a) Find the surface area of S Note that the surface S consists of a portion of the paraboloid z = x2 y2 and a portion of the plane z = 4 Solution Let S1 be the part of the paraboloid z = x2 y2 that liesThe part of the paraboloid z = 1 – x 2 – y 2 that lies above the plane z = –2 more_vert Find the area of the surface 5 The part of the paraboloid z = 1 – x 2 – y 2 that lies above the plane z = –2 close Start your trial now!132 13 MULTIPLE INTEGRALS Example Find the volume of the solid that lies under the paraboloid z = x2 y2, above the xyplane, and inside the cylinder x2 y2 = 2x Completing the square, (x 1)2 y2 = 1 is the shadow of the cylinder in the xyplane Changing to polar coordinates, the shadow of the cylinder is r2 = 2rcos or r = 2cos , so
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Example Find the volume of the solid D bounded by the paraboloid S z = 25−x2 −y2 and the xyplane Solution The paraboloid S z = 25 − x2 − y2 intersect the xyplane p z = 0 in the curve C 0 = 25−x2 −y2, which is a circle x2 y2 = 52 So the shadow R of the solid D after projecting onto xyplane is given by the circular disc R = {(x,y) x2 y2 ≤ 52}, in polar coordinates is Find the volume of the region that lies under the paraboloid \(z = x^2 y^2\) and above the triangle enclosed by the lines \(y = x, \, x = 0\), and \(x y = 2\) in the \(xy\)plane Solution First examine the region over which we need to set up the double integral and the accompanying paraboloidFind the volume below the paraboloid z =1 x^2 y^2 and above the xyplane Expert Answer 100% (3 ratings) Because this problem deals with parabaloid you are going to want to change it to polar First change z = 1 x^2 y^2 into polar This is done by using the fa view the full answer
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