Calculate the volume enclosed by the sphere x^{2}+y^{2}+z^{2}=a^{2}.
Question 5.7.3: Calculate the volume enclosed by the sphere x^2 + y^2 + z^2 ...
In rectangular coordinates, since
\begin{aligned}S=\left\{(x, y, z):-a \leq x \leq a,-\sqrt{a^{2}-x^{2}} \leq y \leq \sqrt{a^{2}-x^{2}}\right., -\sqrt{a^{2}-x^{2}-y^{2}} \leq z \leq \sqrt{\left.a^{2}-x^{2}-y^{2}\right\}}\end{aligned}we can write the volume as
V=\int_{-a}^{a} \int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}} \int_{-\sqrt{a^{2}-x^{2}-y^{2}}}^{\sqrt{a^{2}-x^{2}-y^{2}}} d z d y d xThis expression is very tedious to calculate. So instead, we note that the region enclosed by a sphere can be represented in spherical coordinates as
S=\{(\rho, \theta, \varphi): 0 \leq \rho \leq a, 0 \leq \theta \leq 2 \pi, 0 \leq \varphi \leq \pi\} .Thus
\begin{aligned}V &=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{a} \rho^{2} \sin \varphi d \rho d \varphi d \theta=\int_{0}^{2 \pi} \int_{0}^{\pi} \frac{a^{3}}{3} \sin \varphi d \varphi d \theta \\&=\frac{a^{3}}{3} \int_{0}^{2 \pi}\left\{-\left.\cos \varphi\right|_{0} ^{\pi}\right\} d \theta=\frac{2 a^{3}}{3} \int_{0}^{2 \pi} d \theta=\frac{4 \pi a^{3}}{3}\end{aligned}Related Answered Questions
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