Question

a)Find the area of the surface. The part of the surface z = 5 + 2x + 2y2that lies above the triangle with vertices (0, 0), (0,1), (2, 1).
b)series ne^-n^2 what is the limit and the process

Answer 1

b)
 
 $T_n = n * e^{-n^2},\ \ as\ n\ tends\ to\ 0.\\\\ \frac{T_{n+1}}{T_n}=\frac{(n+1)e^{-(n+1)^2}}{n*e^{-n^2}}=(1+\frac{1}{n})e^{-(n+1)^2+n^2}\\\\ =(1+\frac{1}{n})e^{-2n-1}=\frac{1}{e}(1+\frac{1}{n})(\frac{1}{e^{n}})^2,\\\\ as\ n\ tends\ to\ \infty,\ \ 1/e^n\ tends\ to\ 0.,\ \ \ 1/n\ tends\ to\ 0.\\\\Hence,\ T_{\infty}=0$

a)

$Let\ O=(0,0), A=(0,1)\ and\ B=(2,1)\\\\z=5+2x+2y^2,\ \ between\ lines\ OA: x = 0\ and\ OB:\ x = 2 y \\\\Surface\ area=$

$= \int\limits^{1}_{y=0} { \int\limits^{2y}_{x=0} {\sqrt{1+2^2+(4y)^2} \, dx } \, dy\\\\ = \int\limits^{1}_{y=0} { {\sqrt{1+2^2+(4y)^2} * [ \int\limits^{2y}_{x=0} {1} \, dx ] } \, dy$

$= \int\limits^{1}_{y=0} {2y {\sqrt{1+2^2+(4y)^2} } \, dy \\\\ = \int\limits^{1}_{y=0} {2y {\sqrt{5+16y^2} } \, dy$

$= \frac{1}{24} [ {\sqrt{5+16y^2} ]_0^1\\\\ = \frac{1}{24}(21^{3/2}-5^{3/2})$