Explanation Definition used: The parametric description of the cone is r ( u , v ) = 〈 a v h cos u , a v h sin u , v 〉 with the interval 0 ≤ u ≤ 2 π and 0 ≤ v ≤ h . Definition used: Surface Integral of Scalar-Valued Functions The surface integral of f over the smooth surface S is defined as ∬ S f ( x , y , z ) d S = ∬ R f ( x ( u , v ) , y ( u , v ) , z ( u , v ) ) | t u × t v | d A , where t u = ∂ r ∂ u and t v = ∂ r ∂ v . Calculation: Obtain the parametric representation of r ( u , v ) . Define the parameters u = θ and v = z . The parametric description of the cone is r ( u , v ) = 〈 a v h cos u , a v h sin u , v 〉 with the interval 0 ≤ u ≤ 2 π and 0 ≤ v ≤ h . Differentiate r ( u , v ) with respect to u to get t u = 〈 − a v h sin u , a v h cos u , 0 〉 . Differentiate r ( u , v ) with respect to v to get t v = 〈 a h cos u , a h sin u , 1 〉 . Obtain the value of | t u × t v | . t u × t v = | i ^ j ^ k ^ − a v h sin u a v h cos u 0 a h cos u a h sin u 1 | = { i ^ ( a v h cos u − 0 ) − j ^ ( − a v h sin u − 0 ) + k ^ ( ( − a v h sin u ) ( a h sin u ) − ( a v h cos u ) ( a h cos u ) ) } = ( a v h cos u ) i ^ + ( a v h sin u ) j ^ + ( − a 2 v h 2 sin 2 u − a 2 v h 2 cos 2 u ) k ^ = ( a v h cos u ) i ^ + ( a v h sin u ) j ^ − ( a 2 v h 2 sin 2 u + a 2 v h 2 cos 2 u ) k ^ Further simplify the above expression

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Author: Briggs

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Chapter 14.6, Problem 4E

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The way to compute surface integral of a scalar-valued function over a cone.

Solution Summary: The author explains how to compute the surface integral of a scalar-valued function over the smooth surface S.

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