Mathematics, 25.07.2019 03:10 Horse0809
Compute the surface integral over the given oriented surface: f=y3i+8j−xk, portion of the plane x+y+z=1 in the octant x, y,z≥0, downward-pointing normal
Answers: 2
; in the 1st octant, 
, and:the oriented surface 
![= \int \int_R F. \Big( \dfrac{\partial z}{\partial x}i + \dfrac{\partial z}{\partial y}j - k) dxdy \\ \\ \\ = \int^1_{ x=0} \int ^{1-x}_{y=0} ( y^3i + 8j -xk)*(-i-j-k) dx dy \\ \\ \\ =\int^1_{ x=0} \int ^{1-x}_{y=0} (-y^3 -8+x) dxdy \\ \\ \\ = \int^1_{ x=0} \Big( \int ^{1-x}_{y=0} \Big(x-y^3 -8\Big) dy \Big) dx \\ \\ \\ = \int^1_{ x=0} \Bigg [xy - \dfrac{y^4}{4}-8y \Bigg]^{1-x}_{y=0} \ dx \\ \\ \\](/tpl/images/0129/6135/8afb5.png)
![= \int^1_{ x=0} \Bigg [x(1-x) - \dfrac{1}{4}(1-x)^{4} - 8(1-x) \Bigg ] dx \\ \\ \\ = \int^1_{ x=0} \Bigg [(x-x^2) - \dfrac{1}{4}(x-1)^4+8(x-1)\Bigg] dx](/tpl/images/0129/6135/c73eb.png)
![= \Bigg [ \dfrac{x^2}{2}-\dfrac{x^3}{3} - \dfrac{1}{4} \dfrac{(x-1)^5}{5}+(x-1)^2\Bigg] ^1_0](/tpl/images/0129/6135/8fe4b.png)
![= \Bigg [ \dfrac{1}{2}-\dfrac{1}{3} -0+0-0+0+ \dfrac{1}{4}\times \dfrac{(-1)^5}{5}-(0-1)^2\Bigg]](/tpl/images/0129/6135/2e5f6.png)
![= \Bigg [ \dfrac{1}{2}-\dfrac{1}{3} - \dfrac{1}{20}-1\Bigg]](/tpl/images/0129/6135/1f4a8.png)
![= \Bigg [ \dfrac{30-20-3-60}{60}\Bigg]](/tpl/images/0129/6135/e44df.png)
![= \Bigg [ \dfrac{-53}{60}\Bigg]](/tpl/images/0129/6135/1ca33.png)
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Compute the surface integral over the given oriented surface: f=y3i+8j−xk, portion of the plane x+y...
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