Express the volume of the region enclosed between the surface z = 2 - (x^2 + y^2

Question

Express the volume of the region enclosed between the surface z = 2 - (x^2 + y^2) and z = 6 - 2(x^2 + y^2) as a triple integral in cylindrical coordinates. Then evaluate your integral to find the volume. Figure 1: The surfaces z = 2 - (x^2 + y^2) and z = 6 - 2(x^2 + y^2). Their intersection is dashed. A solid spherical ball centered at the origin with radius R > 0 has varying density given by delta = x^2 + y^2 + z^2 grams per cubic meter. Express the mass of the ball as a triple integral in spherical coordinates. Then evaluate your integral to find its mass. Give you answer with appropriate units. Let G be the region bounded below by the cone phi = pi/6 and bounded above by the plane z = 1. Express the volume of G as a triple integral using the following orders of integration... ... rectangular dz dy dx ... cylindrical dz dr d theta ... cylindrical dr dz d theta ... spherical d rho d phi d theta ... spherical d phi d rho d theta Include any sketches you consider relevant to show your reasoning. Then, using a computer, evaluate each integral to see that they all agree; if they don't, something is wrong. Part(e) will be the most challenging and requires the sum of two integrals to compute. For some guidance, see the following animations which might not be to scale: how to find phi when rho is small: https://msoe.box.com/rhosmall how to find phi when rho is large: https://msoe.box.come/rhobig

Explanation / Answer

in cylinderical coordinate system :

x = rcost

y = rsint

and z = z

and r^2 = x^2+y^2

the range for z is z E [2-r^2 , 6-2r^2]

and the range for t is : t E [0 , 2pi]

and for r equate the given two equations

=> 2-r^2 = 6-2r^2

=> r^2 = 4

or r = 2

=> r E [0,2]

hene the voulme is :

V = integral (r=0 to 2) integral (t=0 to 2pi) integral (z = 2-r^2, ,6-2r^2) [1.dV]

V = integral (r=0 to 2) integral (t=0 to 2pi) integral (z = 2-r^2, ,6-2r^2) [1.]rdzdtdr

V = integral (r=0 to 2) integral (t=0 to 2pi) [rz] (z = 2-r^2, to6-2r^2) dtdr

V = integral (r=0 to 2) integral (t=0 to 2pi) [r(4-r^2)]dtdr

V = integral (r=0 to 2) integral (t=0 to 2pi) [4r - r^3]dtdr

V = integral (r=0 to 2) [4r - r^3]*t (t = 0 to 2pi) dr

V = integral (r=0 to 2) [4r - r^3][2pi-0] dr

V = 2pi*integral (r=0 to 2) [4r - r^3]dt

V = 2pi[2r^2-r^4/4] (r = 0 to 2)

V = 2pi[4]

V = 8*pi units cube , <======== the volume

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