Write a double integral that represents the surface area $ofz=f(x,y)$ that lies above the region $R.$ Use a computer algebra system $to$ evaluate the double integral.

$f(x,y)=4x+y^2$

$R$ : triangle with vertices $(0,0),(8,0),(8,8)$

How $do$ you find the area $of$ a region

$R=\{(r,)$:$g()rh(),\}$

$A={\displaystyle \underset{}{\overset{}{}}}\frac{1}{2}(h({)}^2-g({)}^2)d,$ where

$R=\{(,)$:$g()rh(),\}$

$A={\displaystyle \underset{}{\overset{}{}}}{\displaystyle \underset{g()}{\overset{h()}{}}}rdrd,$ where

$R=\{(r,)$:$g()rh(),\}$

$A={\displaystyle \underset{}{\overset{}{}}}{\displaystyle \underset{g()}{\overset{h()}{}}}drd,$ where

$R=\{(r,)$:$g()rh(),\}$

$(option1$ was incorrect$)$

Change the order $of$ integration $in$ the integral

${\displaystyle \underset{0}{\overset{1}{}}}{\displaystyle \underset{y^2}{\overset{\sqrt[\phantom{2}]{y}}{}}}f(x,y)dxdy$

after changing the order Match the correct answers below for the integral

${\displaystyle \underset{c}{\overset{d}{}}}{\displaystyle \underset{a}{\overset{b}{}}}f(x,y)dydx$

Note: $(x^{\frac{1}{2}}=\sqrt[\phantom{2}]{x},x^2=x^2)$

$a$

$b$

$c$

$d$