Evaluate $ \iint_{S} f(x, y, z) d S $, Where $f(x,y,z)= {x^2}+{y^2}+{z^2}$ and S is given by $ z=x+y, x^{2}+y^{2} \leq 100 $
A) 10,000
B) $10,000 \pi$
C) $10,000 \sqrt { 3 } \pi$
D) $10,000 \sqrt { 2 } \pi$
E) $10,000 \sqrt { 3 }$

Question

Quizplus · Accepted Answer

The surface integral $ \iint_{S} f(x, y, z) d S $ where $f(x,y,z)= {x^2}+{y^2}+{z^2}$ and $S$ is defined by $z=x+y$ with $x^{2}+y^{2} \leq 100$, can be solved by converting to cylindrical coordinates due to the circular symmetry in the $x-y$ plane. In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, and $z = r\cos\theta + r\sin\theta$. The constraint $x^{2}+y^{2} \leq 100$ becomes $r^2 \leq 100$, implying $r \leq 10$.The function $f(x,y,z)$ in cylindrical coordinates becomes $r^2 + (r\cos\theta + r\sin\theta)^2$. The surface element $dS$ in cylindrical coordinates on a surface $z = g(x,y)$ is given by $dS = \sqrt{1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2}r dr d\theta$. For $z = x + y$, $\frac{\partial z}{\partial x} = 1$ and $\frac{\partial z}{\partial y} = 1$, so $dS = \sqrt{3}r dr d\theta$.The integral becomes $\iint_{S} (r^2 + (r\cos\theta + r\sin\theta)^2) \sqrt{3}r dr d\theta$, which simplifies to $\sqrt{3}\int_{0}^{2\pi}d\theta \int_{0}^{10} (2r^2) r dr$, because $r\cos\theta + r\sin\theta$^2 simplifies to $r^2$ when integrated over $0$ to $2\pi$.Evaluating the integral gives $\sqrt{3} \cdot 2\pi \cdot \frac{1}{4} \cdot 10^4 = 10,000\sqrt{3}\pi$, matching choice (C).

Evaluate ∬Sf(x,y,z)dS \iint_{S} f(x, y, z) d S ∬S​f(x,y,z)dS , Where f(x,y,z)=x2+y2+z2f(x,y,z)= {x^2}+{y^2}+{z^2}f(x,y,z)=x2+y2+z2

Evaluate $\iint_{S} f(x, y, z) d S$ , Where $f(x,y,z)= {x^2}+{y^2}+{z^2}$