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?