Chapter 14: Q. 57 (page 1121)

Show that reversing the orientation of a surface S reverses the sign of $\int_{S} F (x, y, z) \cdot n d S$

Short Answer

Expert verified

Ans: It is proved that reversing the orientation of a surface S reverses the sign of $\int_{S} F (x, y, z) \cdot n d S$

Step by step solution

Step 1. Given information.

given, $\int_{S} F (x, y, z) \cdot n d S$

Step 2. The objective is to show that reversing the orientation of a surface S reverses the sign of the above integral.

Reversing the orientation of a surface S means substituting -n instead of n in the above integral.

Substitute -n instead of n in the above integral $\int_{S} F (x, y, z) \cdot n d S$ , then the above integral will be,

$\int_{S} F (x, y, z) \cdot (- n) d S = - \int_{S} F (x, y, z) \cdot n d S$

Hence, the sign of $\int_{S} F (x, y, z) \cdot n d S$ is reversed, if you substitute -n in this integral.

This shows that reversing the orientation of a surface S reverses the sign of the above integral.

Hence Proved.

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Show that reversing the orientation of a surface S reverses the sign of $\int_{S} F (x, y, z) \cdot n d S$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. The objective is to show that reversing the orientation of a surface S reverses the sign of the above integral.

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Show that reversing the orientation of a surface S reverses the sign of ∫S F(x,y,z)⋅ndS

Short Answer

Step by step solution

Step 1. Given information.

Step 2. The objective is to show that reversing the orientation of a surface S reverses the sign of the above integral.

One App. One Place for Learning.

Most popular questions from this chapter

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Show that reversing the orientation of a surface S reverses the sign of $\int_{S} F (x, y, z) \cdot n d S$