1- Find the area using double integration. Y=-X^2 , X-Y=2 2- Find the area using

Question

1- Find the area using double integration. Y=-X^2 , X-Y=2

2- Find the area using double integration r=3 Sin(2*theta)

Explanation / Answer

Due to symmetry, x (centroid) = 0 I divided the figure into 4 sections and concentrate on the regions where x >0 . The wedge between the curves y = -x^2 , line x =y +2 and line y =0 which I called A1 Find the area A1 using double integral It is double integral (1 , x = sqrt ( -y) to y+2 ) , y from -1 to 0 ) A1 = 5/6 Find the centroid of this area. The quadrilateral in the first quadrant formed by line x =0 , y =0 , x =y +2 and y =2 Call this area A2 and find centroid of A2 using geometry or double integral . I found A1 = 5/6 , A2 = 6 So A1 + A2 = 5/6 + 6 = 41 /6 Edit : forget about 4 sections above . Finally, I found and verified that your answer is correct As before, I will concentrate on the right half plane. the area is on the right side and is found using double integral as double integral [ ( 1 , y from - x^2 to 2 ) , x from 0 to 1 ] + double integral [ ( 1 , y from x -2 to 2 ) , x from 1 to 4 ) ] A= 41/6 ( as verified above) The way I approached this problem is as follows: Due to symmetry, x (centroid) = 0 The wedge between the curves y = -x^2 , line x =y +2 and line y =0 which I called A1 Find the area A1 using double integral It is double integral (1 , x = sqrt ( -y) to y+2 ) , y from -1 to 0 ) A1 = 5/6 Find the centroid of this area. The quadrilateral in the first quadrant formed by line x =0 , y =0 , x =y +2 and y =2 Call this area A2 and find centroid of A2 using geometry or double integral . I found A1 = 5/6 , A2 = 6 So A1 + A2 = 5/6 + 6 = 41 /6 Finally, I found and verified that your answer is correct EDIT2 : As before, I will concentrate on the right half plane. the area is on the right side and is found using double integral as double integral [ ( 1 , y from - x^2 to 2 ) , x from 0 to 1 ] + double integral [ ( 1 , y from x -2 to 2 ) , x from 1 to 4 ) ] Now I need to find this double integral quantity I call p1 p1 = { double integral [ ( x , y from - x^2 to 2 ) , x from 0 to 1 ] + double integral [ ( x , y from x -2 to 2 ) , x from 1 to 4 ) ] } p1 = 41/4 and this quantity I call p2 p2 = { double integral [ ( y , y from - x^2 to 2 ) , x from 0 to 1 ] + double integral [ ( y , y from x -2 to 2 ) , x from 1 to 4 ) ] } p2 = 32 /5 So the centroid of the area in the right side plane is x( centroid) = p1 /A = 3/2 y( centroid) = p2 /A = 192/205 Finally, we will have 2 points P1 ( 3/2 , 192 /205 ) and P2 ( -3/2, 192/205 ) Therefore the centroid of the system is at x ( centroid system) = 0 ( due to symmetry ) y ( centroid system ) = 192/ 205

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