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IrinaVladis [17]
2 years ago
9

Find the average value of the function f(x, y, z) = 5x2z 5y2z over the region enclosed by the paraboloid z = 4 − x2 − y2 and the

plane z = 0.
Mathematics
1 answer:
S_A_V [24]2 years ago
3 0
The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin. 

<span>Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. </span>

<span>If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV </span>

<span>∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz </span>

<span>= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 </span>


<span>∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz </span>

<span>= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ </span>

<span>= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ </span>

<span>= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 </span>

<span>∴ F = 10935π/8 ÷ 81π/2 = 135/4</span>
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SOVA2 [1]

y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}

The arc length of the curve is

\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx

which has a value of about 5.99086.

Let f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}. Split up the interval of integration into 10 subintervals,

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The left and right endpoints are given respectively by the sequences,

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These subintervals have midpoints given by

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Over each subinterval, we approximate f(x) with the quadratic polynomial

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\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086

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