Find the area of the portion of the surface $f ( x , y ) = 7 + \frac { 2 } { 3 } y ^ { \frac { 3 } { 2 } }$
that lies above the region $R = \{ ( x , y ) : 0 \leq x \leq 4,0 \leq y \leq 4 - x \}$. Round your answer to two decimal places.
A) $0.05$
B) $2.18$
C) $11.97$
D) $1.22$
E) $12.51$

Question

Quizplus · Accepted Answer

The area of the surface can be found by integrating the function over the given region R. The function $f(x, y) = 7 + \frac{2}{3}y^{\frac{3}{2}}$ does not depend on $x$, so the integration over $x$ is straightforward and only scales the result of integrating over $y$. The region $R$ is defined for $0 \leq x \leq 4$ and $0 \leq y \leq 4 - x$, which means we integrate $y$ from 0 to $4-x$ and then $x$ from 0 to 4. The correct calculation involves setting up and evaluating the double integral of the surface function over the region $R$, which results in an area of approximately $11.97$, making option C the correct answer.

Find the Area of the Portion of the Surface f(x,y)=7+23y32f ( x , y ) = 7 + \frac { 2 } { 3 } y ^ { \frac { 3 } { 2 } }f(x,y)=7+32​y23​

Find the Area of the Portion of the Surface $f ( x , y ) = 7 + \frac { 2 } { 3 } y ^ { \frac { 3 } { 2 } }$