Triple integral with spherical coordinates?

Please write your answer in a legible manner.

Thank you :)!

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2 Answers

  • kb
    Lv 7
    2 years ago
    Favorite Answer

    Following the hint:

    Note that z ≤ 0 iff π/2 ≤ φ ≤ π.

    So, ∫∫∫E ze^((x² + y² + z²)⁶) dV

    = ∫(θ = 0 to 2π) ∫(φ = π/2 to π) ∫(ρ = 0 to 1) (ρ cos φ) e^((ρ²)⁶) * (ρ² sin φ dρ dφ dθ), via spherical coordinates

    = 2π ∫(φ = π/2 to π) sin φ cos φ dφ * ∫(ρ = 0 to 1) ρ³ e^(ρ¹²) dρ

    = 2π * (1/2)sin²φ {for φ = π/2 to π} * ∫(ρ = 0 to 1) ρ³ e^(ρ¹²) dρ

    = -π ∫(ρ = 0 to 1) ρ³ e^(ρ¹²) dρ

    < 0, since ρ³e^(ρ¹²) > 0 on (0, 1] ==> ∫(ρ = 0 to 1) ρ³ e^(ρ¹²) dρ > 0.


    Alternately, one can obtain this directly from the triple integral by noting that e^((x² + y² + z²)⁶) > 0 for all (x,y,z), while we are given z ≤ 0. Then, the integrand for the triple integral is non-positive for all (x,y,z) in the region of integration.

    I hope this helps!

    • Sara2 years agoReport

      Thank you, your explanation is very clear and your help is much appreciated!

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  • rotchm
    Lv 7
    2 years ago

    1-looks like you are trying to cheat. BUSTED

    2- Hint: You don't even need to evaluate the integral(s) to deduce that it is all < 0.

    Now that someone has spoiled you the answer, note that the integrand is always negative on your domain. Hence its integral will be negative.

    • ...Show all comments
    • rotchm
      Lv 7
      2 years agoReport

      I didnt accuse you. But I get it that you (mis)interpreted and that you took it personal when I said " you are trying to cheat". The "you" in there did not mean YOU personally, although it might seem that way. 99% of people misinterpret what they read.

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