Question

What is the value of integral \(∭_Re^{{(x^2+y^2+z^2)}^{\frac{3}{2}}} \,dx \,dy \,dz \) where R is the region given by  x^2+y^2+z^2≤1?

(a) \(\frac{4π(e-1)}{3}\)

(b) \(\frac{4π(e^3-1)}{3}\)

(c) \(\frac{4π(e^2+1)}{3}\)

(d) \(\frac{8π(e+1)}{3}\)

The question was asked in examination.

My doubt is from Change of Variables In a Triple Integral in portion Multiple Integrals of Engineering Mathematics

Answer 1

The correct choice is (a) \(\frac{4π(e-1)}{3}\)

To explain I would say: It can be noticed that R is the region bounded by sphere from the equation x^2+y^2+z^2≤1 thus we are using spherical coordinate to solve this problem

i.e clearly radius r varies from 0 to 1, θ varies from 0 to π & ∅ varies from 0 to 2π

thus the given integral changes to \(\displaystyle∭_{R^*} e^{{r}^{{2}^{1.5}}}\)  r^2  sin⁡θ  dr dθ d∅

\(e^{{r}^{{2}^{1.5}}}\) is obtained by substituting x = r sin θ cos ∅, y = r sin θ sin ∅, z=r cos θ & hence solving the same, now substituting R^* we get

\(\displaystyle\int_ 0^{2π} \int_ 0^π \int_ 0^1 e^{{r}^{{2}^{1.5}}} r^2 \, sin⁡θ \,dr \,dθ \,d∅ = \int_ 0^{2π} d∅ \int_ 0^π sin⁡θ  \,dθ \int_ 0^1 r^2 e^{{r}^{3}} \,dr\)

\(2π*\Big[-cos⁡θ\Big]_0^π * \frac{1}{3} \Big[e^{{r}^{3}}\Big]_{r^3=0}^{r^3=1}=\frac{4π(e-1)}{3}.\)