evaluate the triple integral of sqrt(x^2+9y^2+16z^2) over the solid enclosed by the ellipsoid x^2+9y^2+16z^2=1.
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Note that:
x^2 + 9y^2 + 16z^2 = x^2 + (3y)^2 + (4z)^2,
which suggests that we change to a spherical coordinates, slightly modified:
x = p*sinφcosθ, 3y = p*sinφsinθ, 4z = p*cosφ
==> x = p*sinφcosθ, y = (p/3)sinφsinθ, z = (p/4)cosφ.
(The ranges for θ and φ are 0 <= θ <= 2π and 0 <= φ <= π.)
This transformation has jacobian:
J = |. . . .∂x/∂p. . . .∂x/∂φ. . . .∂x/dθ. . . .|
. . . |. . . .∂y/∂p. . . .∂y/∂φ. . . .∂y/dθ. . . .|
. . . |. . . .∂z/∂p. . . .∂z/∂φ. . . .∂z/dθ. . . .|
. . . |. . . . .sinφcosθ. . . . .p*cosφcosθ. . . . .-p*sinφsinθ. . . .|
=. . |. . . .(1/3)sinφcosθ. . .(p/3)cosφsinθ. . . .(p/3)sinφcosθ. .|
. . . |. . . . (1/4)cosφ. . . . . .(p/4)cosφ. . . . . . . . . . 0. . . . . . .|
= (sinφcosθ)[0 - (p^2/12)sinφcos^2θ] - (p*cosφcosθ)[0 - (p^2/12)sinφcosφcosθ] + (-p*sinφsinθ)[(p^2/12)sinφcosφcosθ - (p^2/12)cos^2φsinθ]
= (-p^2/12)sin^2φcos^3θ + (p^3/12)sinφcos^2φcos^2θ - (p^3/12)sin^2φcosφsinθcosθ + (p^3/12)cos^2φsinφsin^2θ
= (-p^2/12)(sin^2φcos^3θ - p*sinφcos^2φcos^2θ + p*sin^2φcosφsinθcosθ - p*cos^2φsinφsin^2θ)
= (-p^2*sinφ/12)(sin^φcos^3θ - p*cos^2φcos^2θ + p*sinφcosφsinθcosθ - p*cos^2φsin^2θ).
I think that this is as clean as this expression can get.
Since p^2 = x^2 + 9y^2 + 16z^2, the integral becomes:
∫∫∫E p(-p^2*sinφ/12)(sin^φcos^3θ - p*cos^2φcos^2θ + p*sinφcosφsinθcosθ - p*cos^2φsin^2θ) dp dθ dφ (from p=0 to 1) (from θ=0 to 2π and φ=0 to π).
This integral can be done, but it'll take some work. Try it!
I hope this helps!
x^2 + 9y^2 + 16z^2 = x^2 + (3y)^2 + (4z)^2,
which suggests that we change to a spherical coordinates, slightly modified:
x = p*sinφcosθ, 3y = p*sinφsinθ, 4z = p*cosφ
==> x = p*sinφcosθ, y = (p/3)sinφsinθ, z = (p/4)cosφ.
(The ranges for θ and φ are 0 <= θ <= 2π and 0 <= φ <= π.)
This transformation has jacobian:
J = |. . . .∂x/∂p. . . .∂x/∂φ. . . .∂x/dθ. . . .|
. . . |. . . .∂y/∂p. . . .∂y/∂φ. . . .∂y/dθ. . . .|
. . . |. . . .∂z/∂p. . . .∂z/∂φ. . . .∂z/dθ. . . .|
. . . |. . . . .sinφcosθ. . . . .p*cosφcosθ. . . . .-p*sinφsinθ. . . .|
=. . |. . . .(1/3)sinφcosθ. . .(p/3)cosφsinθ. . . .(p/3)sinφcosθ. .|
. . . |. . . . (1/4)cosφ. . . . . .(p/4)cosφ. . . . . . . . . . 0. . . . . . .|
= (sinφcosθ)[0 - (p^2/12)sinφcos^2θ] - (p*cosφcosθ)[0 - (p^2/12)sinφcosφcosθ] + (-p*sinφsinθ)[(p^2/12)sinφcosφcosθ - (p^2/12)cos^2φsinθ]
= (-p^2/12)sin^2φcos^3θ + (p^3/12)sinφcos^2φcos^2θ - (p^3/12)sin^2φcosφsinθcosθ + (p^3/12)cos^2φsinφsin^2θ
= (-p^2/12)(sin^2φcos^3θ - p*sinφcos^2φcos^2θ + p*sin^2φcosφsinθcosθ - p*cos^2φsinφsin^2θ)
= (-p^2*sinφ/12)(sin^φcos^3θ - p*cos^2φcos^2θ + p*sinφcosφsinθcosθ - p*cos^2φsin^2θ).
I think that this is as clean as this expression can get.
Since p^2 = x^2 + 9y^2 + 16z^2, the integral becomes:
∫∫∫E p(-p^2*sinφ/12)(sin^φcos^3θ - p*cos^2φcos^2θ + p*sinφcosφsinθcosθ - p*cos^2φsin^2θ) dp dθ dφ (from p=0 to 1) (from θ=0 to 2π and φ=0 to π).
This integral can be done, but it'll take some work. Try it!
I hope this helps!