∫∫∫ e^abs(z)dv

= 2∫e zdz∫dxdy at this time z ∈ [0, 1]

And ∫∫dxdy can be regarded as a slice of different z of the ball.

That is, X 2+Y 2+Z 2 = 1 At every height z, it can be regarded as a circle with x 2+y 2 =1-z 2.

So: ∫∫ dxdy = π (1-z 2)

rule

The range of 2 ∫ ezdz ∫ dxdy = 2π ∫ ezz (1-z2) dz = z is [0, 1].

=2πe^z-2πz^2e^z+4πze^z-4πe^z

Substitute z ∈ [0, 1]

=2πe-2πe+4πe-4πe+4π

=4π

There is a similar example in the fourth edition of advanced mathematics textbook of Tongji edition, but it is a hemisphere.