Answer:
[tex]\iint_{s}^{} y^2 ds = \dfrac {6250\pi}{3}[/tex]
Step-by-step explanation:
Given :
[tex]x^2 +y^2+z^2=100[/tex]
[tex]x^2+y^2=25[/tex]
Solution:
Parameterize the region in spherical co-ordinates,
[tex]\rm r(\theta ,\phi)=\left\{\begin{matrix}x(\theta , \phi ) & = & 10\;cos\theta sin\phi \\ y(\theta , \phi ) & = & 10\;sin\theta sin\phi \\ z(\theta , \phi ) & = & 10\;cos\phi \end{matrix}\right.[/tex]
where, [tex]0\leq \theta \leq 2\pi[/tex] and [tex]o\leq \phi\leq \dfrac {\pi }{3}[/tex] .Then,
[tex]\rm \left \|r_\theta\times r_\phi \right \| = \left \| \left \langle -100cos\theta sin^2\phi , -100sin\theta sin^2\phi,-100sin\phi cos\phi\right \rangle \right \|= 100sin\phi[/tex]
Surface integral,
[tex]\iint_{s}^{} y^2 ds = \iint_{s}^{} y(\theta ,\phi )^2 \left \| r_\theta\times r_\phi \right \|d\theta d\phi[/tex]
[tex]\iint_{s}^{} y^2 ds = 100^2\int_{\phi = 0}^{\phi = \frac{\pi}{3}}\int_{\theta=0}^{\theta =2\pi }sin^2\theta \; sin^3\phi \; d\theta \; d\phi[/tex]
[tex]\iint_{s}^{} y^2 ds = \dfrac {6250\pi}{3}[/tex]
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https://brainly.com/question/2289273?referrer=searchResults
