Find the area of the surface. The part of the plane $$\mathbf { r } ( u , v ) = ( 5 u + 3 v + 4 ) \mathbf { i } + ( 2 u + 3 v + 4 ) \mathbf { j } + ( 4 u + 2 v + 8 ) \mathbf { k }$$ ; $0 \leq u \leq 1$ , $0 \leq v \leq 5$
A) $5 \sqrt { 149 }$
B) $\sqrt { 58 }$
C) $\sqrt { 51 }$
D) $2 \sqrt { 11 }$

Question

Quizplus · Accepted Answer

Explanation:Given $${\bf r}(u,v)=(5u+3v+4){\bf i}+(2u+3v+4){\bf j}+(4u+2v+8){\bf k}$$$${\bf r}_u=(5{\bf i}+2{\bf j}+4{\bf k})$$$${\bf r}_v=(3{\bf i}+3{\bf j}+2{\bf k})$$$${\bf r}_u	imes {\bf r}_v=\begin{vmatrix}{\bf i}&{\bf j}&{\bf k}\5&2&4\3&3&2\end{vmatrix}$$$$=(4-12){\bf i}-(10-12){\bf j}+(15-6){\bf k}$$$$=-8{\bf i}+2{\bf j}+9{\bf k}$$$$|{\bf r}_u	imes {\bf r}_v|=\sqrt{(-8)^2+(2)^2+(9)^2}=\sqrt{64+4+81}=\sqrt{149}$$Area of the surface $$=\iint\limits_D|{\bf r}_u	imes {\bf r}_v|dA$$$$=\int_0^5\int_0^1\sqrt{149}dudv$$$$=\sqrt{149}\int_0^5\int_0^1dudv$$$$=\sqrt{149}\left[\int_0^5dvight]\left[\int_0^1duight]$$$$=\sqrt{149}[v]_0^5[u]_0^1$$$$=\sqrt{149}(5-0)(1-0)$$$$=5\sqrt{149}$$

Find the Area of the Surface r(u,v)=(5u+3v+4)i+(2u+3v+4)j+(4u+2v+8)k\mathbf { r } ( u , v ) = ( 5 u + 3 v + 4 ) \mathbf { i } + ( 2 u + 3 v + 4 ) \mathbf { j } + ( 4 u + 2 v + 8 ) \mathbf { k }r(u,v)=(5u+3v+4)i+(2u+3v+4)j+(4u+2v+8)k

Find the Area of the Surface $\mathbf { r } ( u , v ) = ( 5 u + 3 v + 4 ) \mathbf { i } + ( 2 u + 3 v + 4 ) \mathbf { j } + ( 4 u + 2 v + 8 ) \mathbf { k }$