Posted by jaime on .
find the area of the region for
2y=4sqrt(x) , y=3, and 2y+3x=7

calculus area between three curves 
jaime,
also help with
Find c>0 such that the area of the region enclosed by the parabolas
y=x^2c^2 and y=c^2x^2 is 430. 
calculus area between three curves 
Reiny,
3 graphs:
y = 2√x
y = 3
y = (3/2)x + 7
see:
http://www.wolframalpha.com/input/?i=plot++y+%3D+2√x+%2C+y+%3D+3+%2C+y+%3D+%283%2F2%29x+%2B+7+%2C+2%3Cx%3C3
intersection of y = (3/2)x+7 and y = 2√x
is aprr A(2.54, 3.188)
intersection of y = 2√x and y = 3
is B(9/4, 3)
intersection of y = (3/2)x + 7 and y = 3 is
C(8/3, 3)
So we want the "trianglular" shaped region ABC
From A, draw a perpendicualar to meet y = 3 at P
So APC is a rightangled triangle
area is easy to find
for the other
area = ∫(2√x  3) dx from x = 9/4 to 2/54
looks routine, messy arithmetic. 
2nd question  calculus area between three curves 
Reiny,
nice symmetry, should be easy to see that the xintercepts are (c,0) and (c,0) for both
so just double the area from 0 to c
area = 2∫(c^2  x^2  (x^2  c^2)) dx from 0 to c
= 2∫(2c^2  2x^2) dx from 0 to c
= 2[2c^2 x  (2/3)x^3] from 0 to c
= 2(c^3  (2/3)c^3]
= 2(1/3)c^3
= 2/3 c^3
but 2/3 c^3 = 430
c^3 = 645
c = ∛645 = appr 8.64
check my arithmetic