# Scalcet7 153019 evaluate the double integral no

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SCalcET7 15.3.019. Evaluate the double integral. (No Response) 2 y 2 dA , D is the triangular region with vertices (0, 1), (1, 2), (4, 1) D 2 2 = 1
9. –/1 points SCalcET7 15.3.029. Find the volume of the given solid. Bounded by the cylinders z = 7 x 2 , y = x 2 and the planes z = 0, y = 4 (No Response) Solution or Explanation Click to View Solution
13. 11. 9. HW #19 6/9 10. –/1 points SCalcET7 15.3.030. Find the volume of the given solid. Bounded by the cylinder and the planes in the first octant is odd with respect to x and is odd with respect to y , and the region of integration is symmetric with respect to both x and y , so represents the volume of the solid region under the graph of and above the rectangle D , namely a half circular cylinder with radius a and length 2 b (see the figure) whose volume is Thus y 2 + z 2 = 16 x = 2 y , x = 0, z = 0 (No Response) Solution or Explanation Click to View Solution 11. –/1 points SCalcET7 15.3.067. Use geometry or symmetry, or both, to evaluate the double integral. (No Response) Solution or Explanation Now is odd with respect to x and is odd with respect to y , and the region of integration is symmetric with respect to both x and y , so represents the volume of the solid region under the graph of and above the rectangle D , namely a half circular cylinder with radius a and length 2 b (see the figure) whose volume is Thus y 2 + z 2 = 16 x = 2 y , x = 0, z = 0 ( 4 ax 9 + 8 by 9 + ) dA , D = [− a , a ] × [− b , b ] a 2 x 2 D = 4 + 8 + ( 4 ax 9 + 8 by 9 + ) dA a 2 x 2 D ax 9 dA D by 9 dA D dA . a 2 x 2 D ax 9 by 9 4 = 8 ax 9 dA D by 9 dA = 0. D dA a 2 x 2 D z = a 2 x 2 · π r 2 h = π a 2 (2 b ) = π a 2 b . 1 2 1 2 ( 4 ax 9 + 8 by 9 + ) dA = 0 + 0 + π a 2 b = π a 2 b . a 2 x 2 D
13. 11. 9. HW #19 12. –/1 points SCalcET7 15.3.063. Use geometry or symmetry, or both, to evaluate the double integral. (No Response) x . D
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