Evaluate [triple integral of (8-(x^2)-(y^2))dV], where H is the solid hemisphere

Question

Evaluate [triple integral of (8-(x^2)-(y^2))dV], where H is the solid hemisphere [[(x^2)+(y^2)+(z^2)] is less than or equal to 9], where z is greater than or equal to 0].

Explanation / Answer

1. x+y=1 ==> y=1-x x+y=1 intersects the coordinate axes at the points (0,1) and (1,0) ==> V = ?[0,1] ?[0,1-x] ?[0,vy] dz dy dx = ?[0,1] ?[0,1-x] vy dy dx = ?[0,1] (2/3)y^(3/2) dx [y from 0 to 1-x] = (2/3)*?[0,1] (1-x)^(3/2) dx = -(2/3)*?[0,1] (1-x)^(3/2) d(1-x) = -(2/3)*(2/5)*(1-x)^(5/2) [x from 0 to 1] = 4/15 2. 3x+6y+4z=12 ==> z=(12-3x-6y)/4 The plane 3x+6y+4z=12 intersects the coordinate axes at the points (4,0,0), (0,2,0) and (0,0,3). z=0 ==> 3x+6y=12 ==> x+2y=4 ==> x=4-2y V = ?[0,2] ?[0,4-2y] ?[0,(12-3x-6y)/4] dz dx dy = ?[0,2] ?[0,4-2y] (12-3x-6y)/4 dx dy = ?[0,2] (12x-3x²/2 -6xy)/4 dy [x from 0 to 4-2y] = ?[0,2] [12(4-2y) -3(4-2y)²/2 -6y(4-2y)]/4 dy = ?[0,2] [3(4-2y) -3(2-y)²/2 -3y(2-y)] dy = ?[0,2] (12-6y-6+6y-3y²/2 -6y+3y²) dy = ?[0,2] (6 -6y+3y²/2) dy = 6y -3y²+y³/2 [y from 0 to 2] = 12 -12+4 = 4 The final result is easy to check, since it is a volume of the pyramid with a right triangle for a base (catheti a=4 and b=2) and a height h=3 ==> V= abh/6 = 4*2*3/6 =4

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