surface integral

[logistic_guy]

here is the question

Evaluate the surface integral. $\displaystyle \iint\limits_S (y^2 + z^2)dS$, where $\displaystyle S$ is the portion of the paraboloid $\displaystyle x = 9 - y^2 - z^2$ in front of the $\displaystyle yz$-plane.


my attemb
i know from basic integration $\displaystyle \int dA = A$
why it's wrong to use the same concept for this integration $\displaystyle \iint\limits_S (y^2 + z^2)dS = (y^2 + z^2)S$?

 

[topsquark]

here is the question

Evaluate the surface integral. $\displaystyle \iint\limits_S (y^2 + z^2)dS$, where $\displaystyle S$ is the portion of the paraboloid $\displaystyle x = 9 - y^2 - z^2$ in front of the $\displaystyle yz$-plane.


my attemb
i know from basic integration $\displaystyle \int dA = A$
why it's wrong to use the same concept for this integration $\displaystyle \iint\limits_S (y^2 + z^2)dS = (y^2 + z^2)S$?

Start with this: sketch the bounding surface.

-Dan

 

[logistic_guy]

Start with this: sketch the bounding surface.

-Dan

to proof to you i'm very good in calculus III, i'm able to visualize the sketch without technology

first i visual
$\displaystyle 0 \leq x \leq 9 - y^2 - z^2$
$\displaystyle -3 \leq y \leq 3$
$\displaystyle -3 \leq z \leq 3$

$\displaystyle \iint\limits_S (y^2 + z^2)dS = \int_{0}^{3}\int_{0}^{3}2(y^2 + z^2)\sqrt{1 + 4y^2 + 4z^4} dy dz = 620$, wrong answer

second i visual
$\displaystyle 0 \leq x \leq 9 - y^2 - z^2$
$\displaystyle -\sqrt{9 - z^2} \leq y \leq \sqrt{9 - z^2}$
$\displaystyle -3 \leq z \leq 3$

$\displaystyle \iint\limits_S (y^2 + z^2)dS = \int_{0}^{3}\int_{0}^{\sqrt{9 - z^2}}2(y^2 + z^2)\sqrt{1 + 4y^2 + 4z^4} dy dz = 312$, wrong answer

after a long think i discover my mistake
$\displaystyle \iint\limits_S (y^2 + z^2)dS = \int_{-3}^{3}\int_{0}^{\sqrt{9 - z^2}}2(y^2 + z^2)\sqrt{1 + 4y^2 + 4z^4} dy dz = 625$, correct answer

i use the website wolfram to solve the integration. is there a method to solve it by hand?