"The value of `(x^2(yz)^2)/((xz)^2y^2)(y^2(xz)^2)/((xy)^2z^2)(z^2(xy)^2)/((yz)^2x^2)`is`1`(b) 0 (c) 1 (d) None of these"This integral of a function along a curve C is often written in abbreviated form as ∫ C f ( x, y) d s Example 1621 Compute ∫ C y e x d s where C is the line segment from ( 1, 2) to ( 4, 7) We write the line segment as a vector function r = 1, 2 t 3, 5 , 0 ≤ tThe sphere x2 y 2 z = 16 and outside the cylinder x2 y = 4 Solution The sphere x2 y2 z2 = 16 intersects the xyplane along the circle with equation x 2 y = 16 Since the solid is symmetric about the xyplane, we may compute its total volume as twice the volume of the part that lies above the xyplane, and this
The Divergence Theorem
X^2+y^2+z^2=16
X^2+y^2+z^2=16-Guillemin and Pollack Chapter 1, Section 4, Problem 5I did this to better visualise a problem for a Differential Geometry course at the University of MichigExample 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
Factorize 9 X Y 2 16 X 2y 2 Brainly In
Fx(x,y) = sin(xy)xcos(xy) and fy(x,y) = xcos(xy) and fz = cosz At (2,2,0), fx(2,−2,0) = sin0 2cos0 = 2 and fy(2,−2) = 2cos0 = 2 and fz(2,2,0) = 1 Therefore the tangent plane is 0 = 2(x−2)2(y 2)−z = 2x2y −z 2 Test the function f(x,y) = x4 y3 32x − 27y for local maxima, minima and saddle points (18) Solution Find theFigure 2 Part of the region S bounded by x2z2 = a2 and x2 y2 = a2 for x ≥ 0 Note that the projection of region S1 on the y − z plane, call it R is a a square 0 ≤ y ≤ a, 0 ≤ z ≤ a We break Experts are tested by Chegg as specialists in their subject area We review their content and use your feedback to keep the quality high Transcribed image text Find the volume of the solid inside the sphere x^2 y^2 z^2 = 16, above the
The lower bound z = x 2 y 2 z = x 2 y 2 is the upper half of a cone and the upper bound z = 18 − x 2 − y 2 z = 18 − x 2 − y 2 is the upper half of a sphere Therefore, we have 0 ≤ ρ ≤ 18, 0 ≤ ρ ≤ 18, which is 0 ≤ ρ ≤ 3 2 0 ≤ ρ ≤ 3 2 For the ranges of φ, φ, weDivide yz, the coefficient of the x term, by 2 to get \frac{yz}{2} Then add the square of \frac{yz}{2} to both sides of the equation This step makes the4 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 solid
x2 y2 = 16 Note that we can rewrite this equation as (x −0)2 (y −0)2 = 42 This is in the standard form (x −h)2 (y −k)2 = r2 of a circle with centre (h,k) = (0,0) and radius r = 4 So this is a circle of radius 4 centred at the origin graph {x^2y^2 = 16 10, 10, 5, 5} Answer link§162 DOUBLE INTEGRALS OVER GENERAL REGIONS Z 1x2 0 f(x,y,z) dz dy dx 276 §164 TRIPLE INTEGRALS Extra Example Set up the integral to find the mass of the region E bounded by the parabolic cylinder z = 1 y2 and the planes x z = 1, x = 0, and z = 0, given theConstraint 2g(x,y,z)= x y2z2=4 Using Lagrange multipliers, solve ∇f= λ ∇g and g=4 This gives Continued •2(x3)=2xλ (12) •2(y1)=2yλ (13) •2(z1)=2zλ (14) •x 2y z2=4 (15) •The simplest way to solve these equations is to solve for x, y, and z
Surface Area
Solved The Ellipsoid 4 X 2 2 Y 2 Z 2 16 Intersects The Plane Y 2 In An Ellipse Find Parametric Equations For The Tangent Line To This Ellipse At The Point 1 2 2
(Definition 2 in Section 162);The definition of a double integral (Definition 5 in Section 151) To evaluate the surface integral in Equation 1, we approximate the patch area ∆S ij by the area of an approximating parallelogram in the tangent plane , 11 ( , , ) lim ( ) mn ij ij mn S ij f x y z dS f P S ofSolution 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 WeStack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange
Surfaces Part 2
Solved Use Cylindrical Coordinates Evaluate Iiint E X Y Dv Where E Is The Solid That Lies Between The Cylinders X 2 Y 2 1 And X 2 Y 2 16
(x y) dV where Eis the solid that lies between the cylinders x2 y2 = 1 and x 2 y = 16, above the xyplane, and below the plane z= y 4 Hint Use cylindrical coordinates coordinates2 0 hx;y;z4i(r r ˆ)dˆd = Z 2ˇ 0 Z p 2 0 ˆ 2 p 2 ˆ5 8 dˆd = 2ˇ Z 2 0 ˆ2 2 p 2 ˆ5 8 dˆ = 1 3 1 6 = 1 6 1784 Use Stokes' Theorem to evaluate ZZ S curlFdS when F(x;y;z) = x2y3zi sin(xyz)j xyzk, and Sis the part of the cone y2 = x2 z2 that lies between the planes y= 0 and y= 3, oriented in the direction of the positive yaxisSo we have to solve z z z in terms of x x x and y y y x 2 y 2 z 2 = 16 → z 2 = 16 − x 2 − y 2 → z = ± 16 − x 2 − y 2 x^2y^2z^2=16 \rightarrow z^2 = 16 x^2 y^2 \rightarrow z= \pm\sqrt {16 x^2 y^2 } x 2 y 2 z 2 = 16 → z 2 = 16 − x 2 − y 2 → z
If Z X Iy And X 2 Y 2 16 Then The Range Of Abs Abs X Abs Y Is Youtube
The Divergence Theorem
Z= ˆcos˚ It follows that x2 y2 = ˆ2 sin2 ˚ If we de ne the angle to have the same meaning as in polar coordinates, then we have x= ˆsin˚cos ;;˚) Example To convert the point (x;y;z) = (1;R = {(x,y,z) x2 y2 6 22, x2 y2 6 z 6 4} Solution 2 x 2 y 4 z z = x y The symmetry of the region implies x = 0 and y = 0 (We verify this result later on) We only need to compute z Since z = 1 V ZZZ R z dv, we start computing the total volume V We use cylindrical coordinates V = Z 2π 0 Z 0 Z 4 r2 dz rdr dθ = 2π Z 0 z 4 r2 rdr
Using Geometry Calculate The Volume Of The Solid Under Z Sqrt 16 X 2 Y 2 And Over The Circular Disk X 2 Y 2 Leq 16 Study Com
Consider The Region Above The Xy Plane Inside The Sphere X 2 Y 2 Z 2 16 And Outside The Cylinder X 2 Y 2 4 A Sketch The Region By Hand Or With The
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= 2ˇ 2r2 1 4 r4 = 2ˇ(8 4) = 8ˇ 3 Evaluate Z Z Z T y2dxdydz where T is the tetrahedron in the rst constant bounded by the coordinate planes and1i th1;0;0i 11 Consider the paraboloid z = x2 y2 (a) Compute equations for the traces in the z = 0, z = 1, z = 2, and z = 3Example Find the volume of the solid region above the cone z2 = 3(x2 y2) (z ≥ 0) and below the sphere x 2 y 2 z 2 = 4 Soln The sphere x 2 y 2 z 2 = 4 in spherical coordinates is ρ = 2
If A X 2 Y 2 16 And B 9x 2 25y 2 225 Find N A Intersection B Brainly In
Solved Find The Surface Area Of The Part Of The Hemisphere Chegg Com
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