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User:Plynch22

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This user is an advanced mathematician .
a b {\displaystyle \int \limits _{a}^{b},円} {\displaystyle \int \limits _{a}^{b},円} This user knows the difference between an integral and an antiderivative ; and so should you! {\displaystyle \int ,円} {\displaystyle \int ,円}
This user enjoys studying fractals .
This user tries to do the right thing. If they make a mistake, please let them know .
This user knows the Ultimate Answer.
This user is fascinated by false colour astronomy representations.
C# This user can program in C# .
Java This user can program in Java .

Hello, and thank you for wasting your time visiting my user page. My name is Paul and I will leave my last name a really tough mystery for you to solve.

I have been told that I sometimes over-complicate things. I know FOR A FACT that 2 1 1 {\displaystyle 2-1\neq 1} {\displaystyle 2-1\neq 1}, because the truth is 1 2 ( d d x ( x ) ) d x = 1 {\displaystyle \int _{1}^{2}\left({\frac {d}{dx}}\left(x\right)\right)dx=1} {\displaystyle \int _{1}^{2}\left({\frac {d}{dx}}\left(x\right)\right)dx=1}. You can't always believe what people say about you.

My Problem

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One thing I would really appreciate is if a sadistic Calc 202 teacher could have their students evaluate this problem for me in both exact notation and approximation.

When r ( u , v ) = ( π + cos v 2 e π ) cos u , ( π + cos v 2 e π ) sin u , sin v 2 e π {\displaystyle {\vec {r}}(u,v)=\left\langle \left(\pi +{\frac {\cos v}{\sqrt[{\pi }]{2^{e}}}}\right)\cos u,\left(\pi +{\frac {\cos v}{\sqrt[{\pi }]{2^{e}}}}\right)\sin u,{\frac {\sin v}{\sqrt[{\pi }]{2^{e}}}}\right\rangle } {\displaystyle {\vec {r}}(u,v)=\left\langle \left(\pi +{\frac {\cos v}{\sqrt[{\pi }]{2^{e}}}}\right)\cos u,\left(\pi +{\frac {\cos v}{\sqrt[{\pi }]{2^{e}}}}\right)\sin u,{\frac {\sin v}{\sqrt[{\pi }]{2^{e}}}}\right\rangle }, evaluate

0 5 5 e 4 + sin 2 ( x ) 0 e π 1024 π e 2000 sin e 0 2 π sin 2 x d x 13 0 q = 0 1000 sin 2 q | r v × r u | u v c b a k = 1 t = 0 19 6 t k 2 0 9 sin 2 ( e f ) d f c b a x g g x {\displaystyle \int _{0}^{5}\int _{5}^{e^{4}+\sin ^{2}(x)}\left.{\frac {\int _{0}^{e^{\pi }}\int _{1024{\sqrt[{e}]{\pi }}}^{2000}\int _{\sin e}^{\int _{\int _{0}^{2\pi }\sin ^{2}x\;dx}^{13}\int _{0}^{\sum _{q=0}^{1000}\sin ^{2}q}\left|{\frac {\partial {\vec {r}}}{\partial v}}\times {\frac {\partial {\vec {r}}}{\partial u}}\right|\;\partial u\;\partial v}\left.{\dfrac {\dfrac {\sqrt[{\sqrt[{a}]{b}}]{c}}{\prod _{k=1}^{\sum _{t=0}^{19}6^{t}}k^{2}}}{\int _{0}^{9}\sin ^{2}\left(e^{f}\right)\;df}}\right.\;\partial c\;\partial b\;\partial a}{x^{\sqrt {g}}}}\right.\;\partial g\;\partial x} {\displaystyle \int _{0}^{5}\int _{5}^{e^{4}+\sin ^{2}(x)}\left.{\frac {\int _{0}^{e^{\pi }}\int _{1024{\sqrt[{e}]{\pi }}}^{2000}\int _{\sin e}^{\int _{\int _{0}^{2\pi }\sin ^{2}x\;dx}^{13}\int _{0}^{\sum _{q=0}^{1000}\sin ^{2}q}\left|{\frac {\partial {\vec {r}}}{\partial v}}\times {\frac {\partial {\vec {r}}}{\partial u}}\right|\;\partial u\;\partial v}\left.{\dfrac {\dfrac {\sqrt[{\sqrt[{a}]{b}}]{c}}{\prod _{k=1}^{\sum _{t=0}^{19}6^{t}}k^{2}}}{\int _{0}^{9}\sin ^{2}\left(e^{f}\right)\;df}}\right.\;\partial c\;\partial b\;\partial a}{x^{\sqrt {g}}}}\right.\;\partial g\;\partial x}


Sadistic calc problem currently under construction.

Of course, I also like Boobs and Alcohol and the chemical Δ 9 {\displaystyle \Delta ^{9}} {\displaystyle \Delta ^{9}}-tetrahydrocannabinol.

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