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# Three Problems Involving Multiple Integrals

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Q1a)Find The area of the paraboloid x^2+y^2=z inside the cylinder x^2+y^2=9

b)write a triple integral in cylindrical coordinates for the volume inside the cone

z^2=x^2+y^2 and between the planes z=1 ans z=2

c)Find the moment of inertia of a circular disk (uniform density) about an axis

through its center and perpendicular to the plane of the disk

https://brainmass.com/math/integrals/three-problems-involving-multiple-integrals-427446

## SOLUTION This solution is FREE courtesy of BrainMass!

a) The area is given by the double integral

A = integral_S(dA) = integral_0^{2 pi}(integral_0^3(rho sec(psi(rho)) drho dphi))

where rho = sqrt(x^2 + y^2) is the cylindrical radial coordinate, phi is the azimuthal angle, and psi(rho) is the inclination angle of the paraboloid at a given point (which depends only on rho). Now the equation for the paraboloid in cylindrical coordinates is

z = rho^2

so we have

tan(psi(rho)) = dz/drho = 2 rho

whence

sec(psi(rho)) = sqrt(1 + tan^2(psi(rho))) = sqrt(1 + 4 rho^2).

Thus we have

A = integral_0^{2 pi}(integral_0^3(rho sqrt(1 + 4 rho^2) drho dphi))
= 2 pi integral_0^3(rho sqrt(1 + 4 rho^2) drho.

Next we make the substitution

u = 1 + 4 rho^2

whence

du = 8 rho drho

and

A = pi/4 integral_1^37(u^(1/2) du)
= pi/6 [u^(3/2)]_{u=1}^37
= pi/6[37 sqrt(37) - 1].

b) In cylindrical coordinates the equation of the cone is |z| = rho, so the integration region is characterized by 0 <= rho <= z, 0 <= phi <= 2 pi and 1 <= z <= 2

We also have dV = rho drho dphi dz. Thus the volume is given by

V = integral_1^2(integral_0^{2 pi}(integral_0^z(rho drho dphi dz))).

c) The moment of inertia is given by the surface integral

I = integral_S(rho^2 dm)

where S is the surface of the disk and

dm = sigma dA = sigma rho drho dphi

sigma = M/A = M/(pi R^2) being the (uniform) mass density (mass per unit area) of the disk, where M is the mass of the disk and R is its radius. Thus we have

I = integral_0^{2 pi}(integal_0^R(sigma rho^3 drho dphi))
= M/(pi R^2)(2 pi) integral_0^R(rho^3 drho)
= 2M/R^2 [rho^4/4]_{rho=0}^R
= 2M/R^2 (R^4/4)
= MR^2/2.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!