Question

Find the surface area of the part of the sphere x2+y2+z2=25 that lies above the cone z=x2+y2‾‾‾‾‾‾‾√

Answer 1

Answer:

$A=50(1-1/\sqrt{2})\pi$ square unit

Step-by-step explanation:

We are given that

$x^2+y^2+z^2=25$

$z=\sqrt{x^2+y^2}$

Compare the equation with

$x^2+y^2+z^2=r^2$

We get

$r^2=25\implies r=5$

In spherical coordinate system

$x=rsin\phi cos\theta\implies x=5sin\phi cos\theta$

$y=rsin\phi sin\theta=5sin\phi sin\theta$

$z=rcos\phi=5cos\phi$

$5cos\phi=\sqrt{25sin^2\phi cos^2\theta+25sin^2\phi sin^2\theta}$

$5cos\phi=5\sqrt{sin^2\phi(cos^2\theta+sin^2\theta)}$

$cos\phi=sin\phi$

$tan\phi=\frac{sin\phi}{cos\phi}=1$

$tan\phi =tan(\pi/4)$

$\phi=\pi/4$

$0\leq \phi\leq \pi/4$

$0\leq \theta \leq 2\pi$

Now, the surface area  of the part of the sphere which lies above the given cone

$A=\int_{\theta=0}^{2\pi}\int_{0}^{\pi/4} r^2 sin\phi d\phi d\theta=\int_{\theta=0}^{2\pi}\int_{0}^{\pi/4}25sin\phi d\phi d\theta$

$A=-25\int_{\theta=0}^{2\pi}[cos\phi]^{\pi/4}_{0} d\theta$

$A=-25\int_{\theta=0}^{2\pi}(1/\sqrt{2}-1)d\theta$

$A=-25(1/\sqrt{2}-1)[\theta]^{2\pi}_{0}$

$A=50(1-1/\sqrt{2})\pi$ square unit