# Evaluate ∫ ∫ ∫ √ 1 − X 2 a 2 − Y 2 B 2 − X 2 C 2 Dx Dy Dz Over the Ellipsoid X 2 a 2 + Y 2 B 2 + Z 2 C 2 = 1 . - Applied Mathematics 2

Sum

Evaluate int int int sqrt(1-x^2/a^2-y^2/b^2-x^2/c^2 )dx dy dz over the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2=1.

#### Solution

Ellipsoid : x^2/a^2+y^2/b^2+z^2/c^2

Cartesian coordinates → spherical coordinate system
(𝒙,𝒚,𝒛) → (𝒓,𝜽,∅)

Put 𝒙=𝒂 𝒓𝒔𝒊𝒏 𝜽 𝒄𝒐𝒔 ∅ ,𝒚=𝒃 𝒓 𝒔𝒊𝒏 𝜽 𝒔𝒊𝒏 ∅ ,𝒛=𝒄 𝒓𝒄𝒐𝒔 𝜽

𝒅𝒙 𝒅𝒚 𝒅𝒛=𝒂𝒃𝒄 𝒓𝟐 𝒔𝒊𝒏 𝜽 𝒅𝒓 𝒅𝜽 𝒅∅

therefore x^2/a^2+y^2/b^2+z^2/c^2=r^2

f(x,y,z)=sqrt(1-x^2/a^2-y^2/b^2-x^2/c^2 )=sqrt(1-r^2)=f(r,theta,O/

Limits :
𝟎 ≤ 𝒓 ≤ 𝟏
𝟎 ≤ 𝜽 ≤ pi/2
𝟎 ≤ ∅ ≤ pi/2

therefore  "I" = 8 int int int sqrt(1-r^2) abc  r^2 sin theta drd thetad O/

= 8 int_0^(pi/2) int_0^(pi/2) int_0^r sqrt(1-r^2) abc  r^2 sin theta drd thetad O/

=8abcint_0^(pi/2)sinthetad thetaint_0^(pi/2) dO/int_0^rsqrt(1-r^2)r^2 dr

8abc[-costheta]_0^(pi/2) [O/]_0^(pi/2)int_0^(pi/2)costsin^2tcost dt ---------{ put r = sint}

=8abc(pi/2)(pi/8) …………{usibetaf^n}

therefore"I"=pi^2/2(abc)

Concept: Triple Integration Definition and Evaluation
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