Question

Evaluate the surface integral.
S(x + y + z) dS, S is the parallelogram with parametric equations
x = u + v, y = u − v, z = 1 + 2u + v, 0 ≤ u ≤ 7, 0 ≤ v ≤ 3.

Answer 1

Answer:

$\frac{115\sqrt{14} }{2}$

Step-by-step explanation:

∫∫ ( x + y + z ) ds

where :

x = u + v ,  y = u - v ,  z = 1 + 2u + v ,  

0 ≤ u ≤ 5,  

0 ≤ v ≤ 1

R ( u,v ) = < u+v , u-v , 1 + 2u +v >   next we have to differentiate

attached below is a detailed solution

Answer 2

The surface integral S(x + y + z) dS, where S is the parallelogram is $115\sqrt{14} /2$ and this can be determined by integrating the given expression.

Given :

The surface integra S(x + y + z) dS, S is the parallelogram with parametric equations  x = u + v, y = u − v, z = 1 + 2u + v, 0 ≤ u ≤ 7, 0 ≤ v ≤ 3.

The following calculation can be used in order to determine the surface integral.

$\rm R_u\times R_v = <3,1,-2>$

$\rm ||R_u\times R_v|| = \sqrt{14}$

Now, evaluate the surface integral.

$\rm I = \rm \int\int_s(x+y+z)dr$

$\rm I = \int^5_{u=0}\int^1_{v=0}(u+v+u-v+1+2u+v)||R_u\times R_v||dA$

$\rm I = \int^5_{u=0}\int^1_{v=0}(u+v+u-v+1+2u+v)\sqrt{14} \;du\;dv$

$\rm I = \sqrt{14} \int^5_0(v+4uv+\dfrac{v^2}{2})^1_0\;du$

$\rm I = \sqrt{14} (\frac{3}{2}u+4\frac{u^2}{2})^5_0$

Simplify the above expression.

$\rm I = \dfrac{115\sqrt{14} }{2}$

The surface integral S(x + y + z) dS, where S is the parallelogram is $115\sqrt{14} /2$

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https://brainly.com/question/22008756