Find the area of the region.
2y=3sqrtx, y=5 and 2y+2x=5
ty
intersection points:
y = 5 with 2y = 3√x ---> (100/9 , 5)
y = 5 with 2x+2y = 5 ---> (-2.5, 5)
2y = 3√x with 2x + 2y = 5 ---> 1 , 3/2)
So I see it as
area
= ∫(5 - (5-2x)/2) dx from -2.5 to 1 + ∫((5-(3/2)(x^.5) )dx from 1 to 100/9
= 6.125 + 14.5185
= .....
not sure of the arithmetic again
Finishing Reiny's excellent evaluation, we get
49/8 + 392/27 = 4459/216 = 20.643
Now, if you want to use horizontal strips, there is no reason to divide up the region, since each strip is bounded on the left and right by a single curve.
Our two boundary curves are now
x = 4/9 y^2
x = (5-2y)/2
and we integrate over 3/2 <= y <= 5
∫[3/2,5] 4/9 y^2 - (5-2y)/2 dy
= 4/27 y^3 + 1/2 y^2 - 5/2 y [3/2,5]
= 4459/216 = 20.6435
http://www.wolframalpha.com/input/?i=plot+y+%3D+3%2F2+sqrt%28x%29%2C+2x%2B2y%3D5%2C+y%3D5
To find the area of the region, we first need to determine the boundaries of the region. We can do this by graphing the given equations and finding the points where they intersect.
The first equation is 2y = 3√x. To graph this equation, we can square both sides to get rid of the square root:
(2y)² = (3√x)²
4y² = 9x
Next, let's graph the equation y = 5. This is a horizontal line that passes through the point (0, 5).
Finally, we can graph the equation 2y + 2x = 5. To do this, we can rearrange the equation to solve for y:
2y = 5 - 2x
y = (5 - 2x)/2
Now we have all three equations graphed. We can find the points of intersection by setting the equations equal to each other:
4y² = 9x
5 = (5 - 2x)/2
Simplifying the second equation, we get:
10 = 5 - 2x
2x = -5
x = -5/2
Substituting this x-value back into the first equation:
4y² = 9(-5/2)
4y² = -45/2
y² = -45/8
Since the value of y² is negative, there are no real solutions for y. Therefore, the two curves do not intersect.
Since there are no points of intersection, the region between the curves is not defined. Therefore, the area of the region is 0.