0
691views
Evaluate $\int\int\int \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2}}$ dx dy dz over the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
1 Answer
0
2views

Solution:

Ellipsoid: $\quad \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}$

enter image description here

$(x, y, z) \quad \longrightarrow \quad(r, \theta, \emptyset)$

Put $x=a \ r \sin \theta \cos \emptyset, \ y=b r \sin \theta \sin \emptyset, \ z=c r \cos \theta$

$d x \ d y \ d z=a b c r^{2} \sin \theta \ d r \ d \theta \ d \emptyset$

$\therefore \quad \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=r^{2}$

$f(x, y, z)=\sqrt{1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}}=\sqrt{1-r^{2}}=f(r, \theta, \emptyset)$

$\begin{aligned} \text { Limits: } & 0 \leq r \leq 1 \\ 0 & \leq \theta \leq \frac{\pi}{2} \\ & 0 \leq \emptyset \leq \frac{\pi}{2} \end{aligned}$

$\begin{aligned} \therefore I &=8 \iiint \sqrt{1-r^{2}} a b c r^{2} \sin \theta \ d r \ d \theta \ d \emptyset \\ &=8 \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{r} \sqrt{1-r^{2}} a b c r^{2} \sin \theta \ d r\ d \theta \ d \phi \\ &=8 a b c \int_{0}^{\frac{\pi}{2}} \sin \theta \ d \theta \int_{0}^{\frac{\pi}{2}} d \varnothing \int_{0}^{r} \sqrt{1-r^{2}} r^{2} d r \end{aligned}$

$\begin{aligned} &=8 a b c[-\cos \theta]_{0}^{\frac{\pi}{2}}[\emptyset]_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \cos t \cdot \sin ^{2} t . \cos t d t \\ &=8 \operatorname{abc}\left(\frac{\pi}{2}\right)\left(\frac{\pi}{8}\right) \\ \therefore I &=\frac{\pi^{2}}{4}(a b c) \end{aligned}$

Please log in to add an answer.