{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

17.5 Prel 1-6, Ex 1-2

# 17.5 Prel 1-6, Ex 1-2 - S E C T I O N 17.5 Surface...

This preview shows pages 1–2. Sign up to view the full content.

S E C T I O N 17.5 Surface Integrals of Vector Fields (ET Section 16.5) 1141 Using the Surface Integral over a Graph we have: Area ( S ) = S 1 dS = D 1 + g 2 x + g 2 y d A (1) In parametrizing the surface by φ ( x , y ) = ( x , y , g ( x , y )) , ( x , y ) = D , we have: T x = x = 1 , 0 , g x T y = y = 0 , 1 , g y Hence, n = T x × T y = i j k 1 0 g x 0 1 g y = − g x i g y j + k = − g x , g y , 1 n = g 2 x + g 2 y + 1 n k k There are two adjacent angles between the normal n and the vertical, and the cosines of these angles are opposite numbers. Therefore we take the absolute value of cos φ to obtain a positive value for Area ( S ) . Using the Formula for the cosine of the angle between two vectors we get: | cos φ | = | n · k | n k = | − g x , g y , 1 · 0 , 0 , 1 | 1 + g 2 x + g 2 y · 1 = 1 1 + g 2 x + g 2 y Substituting in (1) we get: Area ( S ) = D d A | cos φ | 17.5 Surface Integrals of Vector Fields (ET Section 16.5) Preliminary Questions 1. Let F be a vector field and ( u , v) a parametrization of a surface S , and set n = T u × T v . Which of the following is the normal component of F ? (a) F · n (b) F · e n SOLUTION The normal component of F is F · e n rather than F · n . 2. The vector surface integral S F · d S is equal to the scalar surface integral of the function (choose the correct answer): (a) F (b) F · n , where n is a normal vector (c) F · e n , where e n is the unit normal vector SOLUTION The vector surface integral S F · d S is defined as the scalar surface integral of the normal component of F on the oriented surface. That is, S F · d S = S ( F · e n ) dS as stated in (c). 3. S F · d S is zero if (choose the correct answer): (a) F is tangent to S at every point. (b) F is perpendicular to S at every point.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern