The vector field is
A = 5 x y x + x y 2 y + 4 z z . \textbf{A}= 5xy\textbf{x} + xy^2\textbf{y} + 4z\textbf{z}. A = 5 x y x + x y 2 y + 4 z z .
The value of the closed-surface integral ∫ ∫ A d S \int\int AdS ∫∫ A d S
Φ = ∬ A d S = ∭ div A d V , div A = ∂ A x ∂ x + ∂ A y ∂ y + ∂ A z ∂ z = = ∂ 5 x y ∂ x + ∂ x y 2 ∂ y + ∂ 4 z ∂ z = = 5 y + 2 x y + 4. ∭ div A d V = ∭ ( 5 y + 2 x y + 4 ) d x d y d z = = ∫ − 1 2 [ ∫ − 2 4 ( ∫ 1 3 ( 5 y + 2 x y + 4 ) d x ) d y ] d z = = ∫ − 1 2 [ ∫ − 2 4 ( ( 5 x y + x 2 y + 4 x ) ∣ 1 3 ) d y ] d z = = ∫ − 1 2 [ ∫ − 2 4 ( 18 y + 8 ) d y ] d z = = ∫ − 1 2 [ ( y 2 + 8 y ) ∣ − 2 4 ] d z = = ∫ − 1 2 60 d z = 60 z ∣ − 1 2 = 180. \Phi=\iint\textbf{A}\text{ d}S=\iiint\text{ div}\textbf{A}\text{ d}V,\\
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\text{ div}\textbf{A}=\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}=\\
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=\frac{\partial 5xy}{\partial x}+\frac{\partial xy^2}{\partial y}+\frac{\partial 4z}{\partial z}=\\
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=5y+2xy+4.\\
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\iiint\text{ div}\textbf{A}\text{ d}V=\iiint(5y+2xy+4)\text{d}x\text{d}y\text{d}z=\\
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=\int_{-1}^2\Bigg[\int_{-2}^4\Bigg(\int^3_1(5y+2xy+4)\text{d}x\Bigg)\text{d}y\Bigg]\text{d}z=\\
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=\int_{-1}^2\Bigg[\int_{-2}^4\Bigg((5xy+x^2y+4x)\bigg|^3_1\Bigg)\text{d}y\Bigg]\text{d}z=\\
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=\int_{-1}^2\Bigg[\int_{-2}^4(18y+8)\text{d}y\Bigg]\text{d}z=\\
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=\int_{-1}^2\Bigg[(y^2+8y)\bigg|^4_{-2}\Bigg]\text{d}z=\\
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=\int_{-1}^260\text{ d}z=60z\bigg|^2_{-1}=180. Φ = ∬ A d S = ∭ div A d V , div A = ∂ x ∂ A x + ∂ y ∂ A y + ∂ z ∂ A z = = ∂ x ∂ 5 x y + ∂ y ∂ x y 2 + ∂ z ∂ 4 z = = 5 y + 2 x y + 4. ∭ div A d V = ∭ ( 5 y + 2 x y + 4 ) d x d y d z = = ∫ − 1 2 [ ∫ − 2 4 ( ∫ 1 3 ( 5 y + 2 x y + 4 ) d x ) d y ] d z = = ∫ − 1 2 [ ∫ − 2 4 ( ( 5 x y + x 2 y + 4 x ) ∣ ∣ 1 3 ) d y ] d z = = ∫ − 1 2 [ ∫ − 2 4 ( 18 y + 8 ) d y ] d z = = ∫ − 1 2 [ ( y 2 + 8 y ) ∣ ∣ − 2 4 ] d z = = ∫ − 1 2 60 d z = 60 z ∣ ∣ − 1 2 = 180.
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