Math, asked by sumavaishnavi9000, 9 months ago

The average value of a function f(x, y, z) over a solid region E is defined to be fave = 1 V(E) E f(x, y, z) dV where V(E) is the volume of E. For instance, if rho is a density function, then rhoave is the average density of E. Find the average value of the function f(x, y, z) = 3x2z + 3y2z over the region enclosed by the paraboloid z = 4 − x2 − y2 and the plane z = 0.

Answers

Answered by vardhanyashwanth
0

Step-by-step explanation:

The volume of EE is

\displaystyle V(E)=\iiint_E\mathrm dVV(E)=∭

E

dV

To compute the integral, convert to cylindrical coordinates:

x=r\cos\thetax=rcosθ

y=r\sin\thetay=rsinθ

z=zz=z

\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz⟹dV=rdrdθdz

\displaystyle V(E)=\int_0^{2\pi}\int_0^3\int_0^{9-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{81\pi}2V(E)=∫

0

0

3

0

9−r

2

rdzdrdθ=

2

81π

Now integrate ff over EE . In cylindrical coordinates, we get

\displaystyle\iiint_E3x^2z+3y^2z\,\mathrm dV=3\int_0^{2\pi}\int_0^3\int_0^{9-r^2}r^3z\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{6561\pi}8∭

E

3x

2

z+3y

2

zdV=3∫

0

0

3

0

9−r

2

r

3

zdzdrdθ=

8

6561π

Then the average value of ff over EE is \dfrac{\frac{6561\pi}8}{\frac{81\pi}2}=\dfrac{81}4

2

81π

8

6561π

=

4

81

.

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