# Bracketrightbigg 1 1 dz 0 integraldisplay 1 1

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Unformatted text preview: bracketrightbigg 1- 1 dz + 0 + integraldisplay 1- 1 bracketleftbigg e y + e- y bracketrightbigg 1- 1 dx = 4 integraldisplay 1- 1 dz + integraldisplay 1- 1 parenleftbigg e + 1 e- 1 e- e parenrightbigg dx = 8 + 0 = 8. We see that, for this example, the Divergence Theorem holds. 3. Since the conditions are satisfied, we can use the Divergence Theorem. Now integraldisplay S F · d S Divergence = Theorem integraldisplay ball div F dV = integraldisplay ball (3 y 2 + 3 x 2 + 3 z 2 ) dV spherical = polars 3 integraldisplay 2 π integraldisplay π integraldisplay 1 ( ρ 2 ) ρ 2 sin ϕ dρdϕ dθ = (3) (2 π ) bracketleftbigg ρ 5 5 bracketrightbigg 1 integraldisplay π sin ϕ dϕ = 6 π 5 bracketleftbigg- cos ϕ bracketrightbigg π = 12 π 5 . 4. Let the vertex of the cone be at the origin and the base in the plane z = h as shown on the right. The boundary of the solid cone C consists of two pieces, the shell S , and the base region D (which has area A ). Let the vector field F be F ( x, y, z ) = ( x, y, z ). Since F ( x, y, z ) points directly away from the origin at any nonzero point ( x, y, z ), we have F · n = 0 on S . On the base, n = e 3 and z = h , so F · n = h . Now div F = 1 + 1 + 1 = 3 so the Divergence Theorem gives 3 V = integraldisplay C div F dV = integraldisplay ∂C F · n dS = integraldisplay S F · n dS + integraldisplay D F · n dS = 0 + h integraldisplay D dS = Ah . x y z z Equal h D S n n Hence V = 1 3 Ah , the usual formula for the volume of a cone when the height and base area are known. 5. We will use the Divergence Theorem to evaluate integraldisplay S F · n dS . To do this we first con- sider the solid region R shown on the right below. Now S is one of 5 surfaces that x y z S b a x y z S b a MATB42H Solutions # 9 page 3 make up the boundary ∂R . Let S 1 lie in the xy –plane, S 2 lie in the yz –plane, S 3 lie in the xz –plane and S 4 lie in the plane y = b . Clearly F · n = 0 on S 1 , S 2 and S 3 . On S 4 , F · n = ( x, y 2 , z ) · e 2 = y 2 = b 2 . Now integraldisplay ∂R F · n dS = integraldisplay S F · n dS + 0 + 0 + 0 + integraldisplay S 4 F · n dS = integraldisplay S F · n dS + b 2 integraldisplay S 4 dS = integraldisplay S F · n dS + b 2 parenleftbigg π a 2 4 parenrightbigg . Using the Divergence Theorem, we have integraldisplay ∂R F · n dS = integraldisplay R div F dV = 2 integraldisplay R (1 + y ) dV = 2 integraldisplay 2 π integraldisplay r integraldisplay b (1 + y ) r dy dr dθ = 2 parenleftbigg b + b 2 2 parenrightbiggparenleftbigg π 2 parenrightbiggbracketleftbigg r 2 2 bracketrightbigg a = π a 2 b 2 + π a 2 b 2 4 ....
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• Winter '10
• EricMoore

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