Find the surface area of the surface S.
-$S$ is the cap cut from the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 16$ by the cone $z = 6 \sqrt { x ^ { 2 } + y ^ { 2 } }$.
A) $16 \left( 1 - \frac { 6 \sqrt { 37 } } { 37 } \right)$
B) $16 \left( 1 - \frac { 6 \sqrt { 37 } } { 37 } \right) \pi$
C) $32 \left( 1 - \frac { 6 \sqrt { 37 } } { 37 } \right) \pi$
D) $16 \left( \frac { 6 \sqrt { 37 } } { 37 } - 1 \right) \pi$

Question

Quizplus · Accepted Answer

Using spherical coordinates, we have
$$\begin{aligned} x&=r\sin	heta\cos\phi \ y&=r\sin	heta\sin\phi\ z&=r\cos	heta\end{aligned}$$
The equation of the sphere is $$r^2=x^2+y^2+z^2=16$$ and the equation of cone is $$z=6\sqrt{x^2+y^2}=6r\sin	heta.$$ 
The intersection of the sphere and cone is where
$$\begin{aligned} r^2&=16\ r\cos	heta&=6r\sin	heta\implies 	an	heta=\frac16\implies \cos	heta=\sqrt{\frac{1}{1+	an^2	heta}}=\frac{\sqrt{37}}{6\sqrt{37}+1}.\end{aligned}$$
Thus, the surface area of the cap is
$$\begin{aligned} S &=\iint_{	ext{Cap}}\sqrt{\left(\frac{\partial z}{\partial x}ight)^2+\left(\frac{\partial z}{\partial y}ight)^2 +1}\,dA\ &=\iint_{	ext{Cap}}\sqrt{\left(\frac{x}{\sqrt{x^2+y^2}}ight)^2+\left(\frac{y}{\sqrt{x^2+y^2}}ight)^2 +1}\,dA\ &=\int_{0}^{2\pi}\int_{0}^{\arctan\frac16}\sqrt{\frac{1}{\cos^2	heta}+1}\cdot 4\sin	heta\cdot r\,drd	heta\ &=32\pi\int_{0}^{\arctan\frac16}\frac{\sqrt{\sin^2	heta+37}}{6\sqrt{37}+1}\,d	heta\ &=32\pi\cdot\frac{1}{6\sqrt{37}+1}\int_{0}^{\arctan\frac16}\sqrt{37\cos^2	heta+1}\,d	heta\ &=32\pi\cdot\frac{1}{6\sqrt{37}+1}\cdot\frac{\sqrt{37}}{2}\int_{0}^{\arctan\frac16}\sqrt{\frac{\cos^2	heta}{\frac{1}{37}}+1}\,d	heta\ &=32\pi\cdot\frac{1}{6\sqrt{37}+1}\cdot\frac{\sqrt{37}}{2}\cdot\frac{37}{6}\left(1-\frac{6\sqrt{37}}{37}ight)\ &= \boxed{32\left(1-\frac{6\sqrt{37}}{37}ight)\pi}.\end{aligned}$$

Find the Surface Area of the Surface S SSS Is the Cap Cut from the Sphere

Find the Surface Area of the Surface S $S$ Is the Cap Cut from the Sphere