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Hw15-5

# Hw15-5 - MATH1920 HOMEWORK 9 SECTION 15.5 0 0 0 0 0 0 the...

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MATH1920 HOMEWORK 9, SECTION 15.5 2 The six iterated triple integrals are respectively Z 1 0 Z 2 0 Z 3 0 dxdydz , Z 1 0 Z 3 0 Z 2 0 dxdzdy , Z 2 0 Z 1 0 Z 3 0 dydxdz , Z 2 0 Z 3 0 Z 1 0 dydzdx , Z 3 0 Z 1 0 Z 2 0 dzdxdy , Z 3 0 Z 2 0 Z 1 0 dzdydx . It can be easily seen that any one of the integrals is equal to 6. 9 Z e 1 Z e 2 1 Z e 3 1 1 xyz dxdydz = Z e 1 Z e 2 1 ln x yz e 3 1 dydz = Z e 1 Z e 2 1 3 yz dydz = Z e 1 3 ln y z e 2 1 dz = Z e 1 6 z dz = [6 ln z ] e 1 = 6 20 Z 7 0 Z 2 0 Z 4 - q 2 0 q r + 1 dpdqdr = Z 7 0 Z 2 0 q p 4 - q 2 r + 1 dqdr = Z 7 0 " - (4 - q 2 ) 3 2 3( r + 1) # 2 0 dr = Z 7 0 8 3( r + 1) dr = 8 3 ln( r + 1) 7 0 = 8 ln 2 21 Let R be the region of integration. (a) The limits along the y -direction of R are given by the surfaces y = x 2 and y + z = 1, i.e. y = 1 - z . So the integral can be rewritten as Z R xz Z 1 - z x 2 dydA xz 1

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2 MATH1920 HOMEWORK 9, SECTION 15.5 where R xz is the projection of R onto the xz -plane, and dA xz is the area differential of the xz -plane. Along the positive z -direction, R xz is bounded below by z = 0 and above by the projection of the curve of intersection of y + z = 1 and y = x 2 onto the xz - plane. The equation of the projection of this curve of intersection is z = 1 - x 2 (which is obtained by eliminating y from the two equations y + z = 1 and y = x 2 ). So if we next integrate the z -direction then its limits in R xz are z = 0 and z = 1 - x 2 . From the picture we can see that the limits of the x -direction in R xz is - 1 and 1, if we integrate
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Hw15-5 - MATH1920 HOMEWORK 9 SECTION 15.5 0 0 0 0 0 0 the...

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