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Chapter 14: Q. 44 (page 1151)

In Exercises 41–44, find the fluxes of the vector fields through the given surfaces in the direction of the outwards-pointing normal vector.

$$F(x,y,z)= \langle x^{2}y^{2}z^{2}, z-x-y,\frac{y}{z+1} \rangle$$, and $$S$$ is the surface of the region bounded by $$z = x^{2}$$, $$x = 0$$, and $$4x +2z = 4$$.

Short Answer

Expert verified

The flux of the given vector field through the given surfaces in the direction of the outwards-pointing normal vector is $$\int \int_{W}\int (2xy^{2}z^{2}-1+yln(z+1))dxdydz$$

Step by step solution

01

Step 1. Given Information

$$F(x,y,z)= \langle x^{2}y^{2}z^{2}, z-x-y,\frac{y}{z+1} \rangle$$, and $$S$$ is the surface of the region bounded by $$z = x^{2}$$, $$x = 0$$, and $$4x +2z = 4$$.

02

Step 2. Explanation

The flux of the vector field can be given as, $$\int \int_{W}\int div F(x,y,z)dV$$ ------(1)

Here, the divergence can be given as, $$divF=2xy^{2}z^{2}-1+yln(z+1)$$ -----(2)

Using (2) in (1), we get the flux as, $$\int \int_{W}\int (2xy^{2}z^{2}-1+yln(z+1))dxdydz$$

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