&nbsp;let us also assign to n the orientation in the direction of increasing x value. then compute surface integral over n of the vector field k on&nbsp;R3&nbsp;plane given byh(x,y,z)=(y,0,z)(x,y,z)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

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&amp;nbsp;let us also assign to n the orientation in the direction of increasing x value. then compute surface integral over n of the vector field k on&amp;nbsp;R3&amp;nbsp;plane given byh(x,y,z)=(y,0,z)(x,y,z)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;

alenahelenash · Accepted Answer

We are given a vector field F in ℝ3 as:F→(x,y,z)=⟨0,0,1⟩And, we are given a surface S in ℝ3 with parameterization:h→(x,y,z)=⟨y,0,z⟩where S is the plane in ℝ3 defined by this parameterization.We are asked to find the surface integral of F→ over the surface S with an orientation in the direction of increasing x values.To compute the surface integral, we can use the formula:∬SF→·dS→=∬RF→(h→(u,v))·(h→u×h→v)dAwhere R is the region in the uv-plane that corresponds to the surface S under the parameterization h→(u,v), and h→u and h→v are the partial derivatives of h→ with respect to u and v.First, we need to find h→u and h→v:h→u=⟨0,0,0⟩h→v=⟨0,1,0⟩Next, we need to find the region R in the uv-plane. The parameterization of S tells us that y can take on any value and z can take on any value, so we have:R={(u,v)∈ℝ2&amp;nbsp;|&amp;nbsp;0≤u≤1,0≤v≤1}Now, we can evaluate the surface integral:∬SF→·dS→=∬RF→(h→(u,v))·(h→u×h→v)dAF→(h→(u,v))=F→(y,0,z)=⟨0,0,1⟩h→u×h→v=⟨0,0,−1⟩Therefore,∬SF→·dS→=∬R⟨0,0,1⟩·⟨0,0,−1⟩dA=∬R0dA=0Thus, the surface integral over S of the vector field F→ is 0.

let us also assign to n the orientation

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