Arrow_forward Buy Find launch Calculus Volume 3First week only $499!The moment with respect to the x axis is M x = ZZ D k ¡ x2 y2 ydA = Z π/2 0 Z 1 0 kr2rsin(θ)rdrdθ Z π/2 0 Z 1 0 kr4 sin(θ) drdθ 1 5 k Z π/2 0 sin(θ) dθ = k 5 Likewise, the moment with respect to the y axis is M y = k/5 Therefore, the center of
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X^2+y^2+z^2=16 x^2+y^2=4
X^2+y^2+z^2=16 x^2+y^2=4- Section 15 Functions of Several Variables In this section we want to go over some of the basic ideas about functions of more than one variable 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 − 4The given sphere is {eq}\displaystyle x^2 y^2 z^2 16 = 10 {/eq} Compare the above equation with general equation of sphere, {eq}\displaystyle x^2 y^2 z^2 2a_1x 2b_1 y 2c_1 z d



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Plane z= 0 and the hemisphere x2y2z2 = 9, bounded above by the hemisphere x2y2z2 = 16, and the planes y= 0 and y= x This would be highly inconvenient to attempt to evaluate in Cartesian coordinates;//googl/JQ8NysFind a Unit Normal Vector to the Sphere x^2 y^2 z^2 = 57 at (4, 4, 5)Evaluate the surface integral ZZ S xyzdS, where Sis the part of the sphere x2 y 2 z 2= 1 that lies above the cone z= p x y Using the spherical
More_vert 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)Surface we have x2 y2 = u2 = z This is the equation for a parabolic bowl centered on the zaxis with vertex at the origin Example 12 Identify the surface with parametric equations ~rx,ϑ) = u~iucos(ϑ)~j usin(ϑ)~k Since x = x, y = xcos(ϑ) and z = xsin(ϑ), at any point on this surface we have y2 z2 = x2 This is the equation for aConsider 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 B Use Polar Coordinates To Find The Volume For more information and source, see on this link https
Describe Sketch And Name These Cylinders And Quadric Surfaces In R 3 A 4 X 2 16 Y 2 Z 2 16 B X 2 4 Y 2 4 Z 2 0 C X 2 Y 3 Z 12 D Y 2 Z 2 9 Study Com For more information and source, see onX^2y^2z^2=1 WolframAlpha Have a question about using WolframAlpha?A sphere is the graph of an equation of the form x 2 y 2 z 2 = p 2 for some real number p The radius of the sphere is p (see the figure below) Ellipsoids are the graphs of equations of the form ax 2 by 2 cz 2 = p 2, where a, b, and c are all positive



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ZdV where E is the portion of the solid sphere x2 y2 z2 ≤ 9 that is inside the cylinder x2 y2 = 1 and above the cone x2 y2 = z2 Figure 5 Soln The top surface is z = u2(x,y) = p 9− x2 − y2 = √ 9− r2 and the bottom surface is z = u1(x,y) = p x2 y2 = r over the region D defined by the intersection of the top (or 4 964 It's not really necessary to use calculus at all!The trace in the z = 1 plane is the ellipse x2 y2 8 = 1, shown below 6 13 Surface 24x 24y2 9z = 35;



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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyThe 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 isNow we save this in



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This is a circle of radius 4 centred at the origin Given x^2y^2=16 Note that we can rewrite this equation as (x0)^2(y0)^2 = 4^2 This is in the standard form (xh)^2(yk)^2 = r^2 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}Express the volume of the solid inside the sphere x 2 y 2 z 2 = 16 and outside the cylinder x 2 y 2 = 4 as triple integrals in cylindrical coordinates and spherical coordinates, respectively close Start your trial now!The cone z = sqrt(x^2 y^2) can be drawn as follows In cylindrical coordinates, the equation of the top half of the cone becomes z = r We draw this from r = 0 to 1, since we will later look at this cone with a sphere of radius 1 > cylinderplot(r,theta,r,r=01,theta=02*Pi);



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132 fishingspree2 said Find the volume inside the sphere x 2 y 2 z z = 16 and outside the cylinder x 2 y 2 = 4 Use polar coordinates The sphere's center lies at the origin The region of integration is the base of the cylinder, the radius 2 xy disk x 2 y 2 = 4 and the two parts of the sphere are given by z = ± 16 − x 2 − y 2Graph x^2y^2=16 x2 − y2 = 16 x 2 y 2 = 16 Find the standard form of the hyperbola Tap for more steps Divide each term by 16 16 to make the right side equal to one x 2 16 − y 2 16 = 16 16 x 2 16 y 2 16 = 16 16 Simplify each term in the equation in order to set the right side equal to 1 1 The standard form of an ellipse orFactorise 4 x 4 9 y 4 6 x 2 y 2 Easy View solution Show that if 2 (a 2 b 2) = (a b) 2, then a = b Easy View solution In each case, state True if the trinomial is a perfect square else state False



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The 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 computeB) Find the area of the piece of the sphere that is inside the cylinderTake 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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It's the equation of sphere The general equation of sphere looks like math(xx_0)^2(yy_0)^2(zz_0)^2=a^2/math Wheremath (x_0,y_0,z_0)/math is the centre of the circle and matha /math is the radious of the circle It's graph looksPlease Subscribe here, thank you!!!Plane x = 1 2 The trace in the x = 1 2 plane is the hyperbola y2 9 z2 4 = 1, shown below For problems 1415, sketch the indicated region 14 The region bounded below by z = p



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16 and outside the cylinder x2 y2 = 4 Solution The sphere x 2 y 2 z 2 = 16 intersects the xyplane in the circle x 2 y 2 = 16, so that the volume in question is computed asA z 2 c 2 y 2 b 2 x 2 a 2 1 b x 2 a 2 y 2 b 2 z 2 c 2 1 c x y 2 b 2 z 2 c 2 d x from MATH 151 at Schoolcraft College Review on Vectors and Space Geometry Let shift the origin of the coordinates to the point $(0,2,0)$ so that the equation of the sphere and cylinder are $$ x^2(y2)^2z^2=16\text{ and } x^2y^2=4,\tag1 $$ respectively



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F(x, y, z) = –y2 i x j z2 k C is the curve of intersection of the plane y z = 2 and the cylinder x2 2 y = 1 (Orient C to be counterclockwise when viewed from above) could be evaluated directly, however, it's easier to use Stokes' Theorem C ∫Fr⋅d Example 1 C ∫Fr⋅d161 Vector Fields This chapter is concerned with applying calculus in the context of vector fields A twodimensional vector field is a function f that maps each point ( x, y) in R 2 to a twodimensional vector u, v , and similarly a threedimensional vector field maps ( x, y, z) to u, v, w Since a vector has no position, we typicallyWe consider the hemi sphere x^2 y^2 z^2 = 16 and the piece of cylinder x^2 y^2 = 4x z greaterthanorequalto 0 a) Do a figure of the cylinder What is its axis?



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Let G be the solid enclosed by the cylinder x^2 y^2 = 4, the sphere x^2 y^2 z^2 = 16, and the plane z = 0 (a) Set up, but do not evaluate, an integral (or integrals) in rectangular coordinates for the volume of G (b) Set up, but do not evaluate, an integral (or integrals) in cylindrical coordinates for the volume of GDetermining the limits in z alone requires breaking up the integral with respect to zZ = x2 y2 lying underneath the plane z = 1, with npointing generally upwards Explain geometrically why your answer is negative 6* Find RR S F·dS, where F = xi yj zk x 2y z2, and S is the surface of Exercise 6B2 6B8 Find RR S FdS, where F= yj and S



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Inside the sphere x^2 y^2 z^2 = 16 and outside the cylinder x^2 y^2 = 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 certificateSolution for x^2y^2z^216=0 equation Simplifying x 2 y 2 z 2 16 = 0 Reorder the terms 16 x 2 y 2 z 2 = 0 Solving 16 x 2 y 2 z 2 = 0 Solving for variable 'x' Move all terms containing x to the left, all other terms to the right Add '16' to each side of the equation 16 x 2 y 2 16 z 2 = 0 16 Reorder the termsFind the surface area of the part of the sphere x^2 y^2 z^2 = 16 inside the cylinder x^2 4x y^2 = 0 Find the surface area of the part of the sphere x 2 y 2 z 2 = 1 6 inside the cylinder x 2



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Answer to Use polar coordinates to find the volume of the given solid Inside the sphere x^2 y^2 z^2 = 16 and outside the cylinder x^2 y^2 =X2 9 y2 36 z2 16 = 1 and below by z = 2 Description in terms of range on z,theny,thenx (b) S is bounded above by x2 16 y2 36 z2 16 = 1 and below by y 6 z 4 = 1 Description in terms of range on z,thenx,theny 2 Each of the following iterated integrals cannot beThe surface on the xyplane is the region between x2 y2 = 4 and x2 y2 = 4, 2 r 4 Therefore the required area is R 2ˇ 0 R 4 2 r p 4r2 1drd 10 (a) Since 2x 2zz x = 0, z x = x z Similarly z y = y z Hence q 1 f2 x f y 2 = q 1 z2 x z y 2 = q 1 x 2 z 2 y z = 2 z The projection of the Son the xyplane is D= f(x;y) x2y2 4g



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Unlock StepbyStep x^2/16y^2/16z^2/16=1 Extended Keyboard ExamplesThe surface area of a function mathz = f(x,y)/math over a region D is math\iint_D \sqrt{1(\frac{\partial z}{\partial x})^2(\frac{\partial z}{\partial y})^2Graph x^2y^2=16 x2 y2 = 16 x 2 y 2 = 16 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard form The variable r r represents the radius of the circle, h h represents the xoffset from



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Stack Exchange network consists of 177 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 ExchangeThe 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 Boost your resume with certification as an expert in up to 15 unique STEM subjects this summer1 8 dt= udu changing the bounds, we get = 1 2 Z 5 1 1 4 (t 1) p t 1 8 dt = 1 64 Z 5 1 t3=2 t1=2 dt 1 64 2 5 t5=2 2 3 t3=2 5 1 = 5 48 p 5 1 240 11 Evaluate RR S x 2z2 dS, where Sis the part of the cone z2 = x2 y between the planes z= 1 and z= 3 The widest point of Sis at the intersection of the cone and the plane z= 3, where x2 y2 = 32 = 9;



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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 SolutionConsider 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 B Use Polar Coordinates To Find The Volume For more information and source, see on this link httpsContact Pro Premium Expert Support »



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Substitute t= 4u2 1;u2 = 1 4 (t 1);Lagrange Multipliers Minimum of f(x, y, z) = x^2 y^2 z^2 subject to x y z 9 = 0 See below Considering the variables x,y,z defining RR^3, this geometrical object is a surface such that for any x obeys y^2z^2= 16 so it is a cylindrical surface centered at the x axis having a circle with radius 4 as transversal cross section



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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 solid



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