# Triple integral with spherical coordinates?

Thank you :)! Relevance

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!

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

• Login to reply the answers
• 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.

• Login to reply the answers