I have this one homework problem and I just can't get it right. Any help would be greatly appreciated! **I know that you are supposed to integrate w/ respect to y because it is a lot easier.
Question: Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
2y=5(x)^1/2 , y=4 and 2y+4x=9
Question: Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
2y=5(x)^1/2 , y=4 and 2y+4x=9
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Yes, integrating with respect to y will be easier. In drawing the three curves,
we see that the region enclosed lies under y = 4 and between x = (4/25)*y^2 and
x = (-1/2)y + (9/4). Now these last two curves intersect when
(4/25)*y^2 = (-1/2)y + (9/4) ---> y*2 + (25/8)y - (225/16) = 0. Using the quadratic
formula and taking the positive solution we get y = 2.5. So the area of the region is
A = integral((4/25)*y^2 - ((-1/2)y + (9/4)) dy (from y = 2.5 to 4), and so
A = ((4/75)*y^3 + (1/4)*y^2 - (9/4)y) (from y = 2.5 to 4) =
(-1.586667) - (-3.2291777) = 1.6425.
we see that the region enclosed lies under y = 4 and between x = (4/25)*y^2 and
x = (-1/2)y + (9/4). Now these last two curves intersect when
(4/25)*y^2 = (-1/2)y + (9/4) ---> y*2 + (25/8)y - (225/16) = 0. Using the quadratic
formula and taking the positive solution we get y = 2.5. So the area of the region is
A = integral((4/25)*y^2 - ((-1/2)y + (9/4)) dy (from y = 2.5 to 4), and so
A = ((4/75)*y^3 + (1/4)*y^2 - (9/4)y) (from y = 2.5 to 4) =
(-1.586667) - (-3.2291777) = 1.6425.
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The first curve is a parabola: 4y² = 25.x
The second curve is a straight line, parallel to the x-axis
The third is a straight line going through (x=0, y =4.5) and (x=2.25, y = 0)
We see that the first and second curve intersect at (x= 64/25 , y =4).
The first and third curve at (x =1, y = 2.5)
The area between the y-axis and the first curve is easily integrated from y = 2.5 tot y =4.
From this we subtract the area between y-axis and the third curve (the line) from y = 2.5 to y = 4
The first area is [4/75 . y³] between 4 and 2.5 is 256/75 - 125/150 = 387/150.
From this we subtract the area between the sloped line and the y-axis. The simplest way is to multiply the width over the y-axis (4 - 2,5 = 1,5) by the value of the x in the middle (at y = 3.75)
There x = 1.5/4 So the area is 387/150 - {1,5 . 1.5/4} = 387/150 - 9/16 = {4 . 387 - 9 . 75} / 1200 = 873 / 1200 = 291/400
Please check my arithmetic, I am clumsy at it.
The second curve is a straight line, parallel to the x-axis
The third is a straight line going through (x=0, y =4.5) and (x=2.25, y = 0)
We see that the first and second curve intersect at (x= 64/25 , y =4).
The first and third curve at (x =1, y = 2.5)
The area between the y-axis and the first curve is easily integrated from y = 2.5 tot y =4.
From this we subtract the area between y-axis and the third curve (the line) from y = 2.5 to y = 4
The first area is [4/75 . y³] between 4 and 2.5 is 256/75 - 125/150 = 387/150.
From this we subtract the area between the sloped line and the y-axis. The simplest way is to multiply the width over the y-axis (4 - 2,5 = 1,5) by the value of the x in the middle (at y = 3.75)
There x = 1.5/4 So the area is 387/150 - {1,5 . 1.5/4} = 387/150 - 9/16 = {4 . 387 - 9 . 75} / 1200 = 873 / 1200 = 291/400
Please check my arithmetic, I am clumsy at it.