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
Find the area of the region.
2y=3sqrtx, y=5 and 2y+2x=5
ty
2 answers
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
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
3 answers