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Posted by **jaime** on Monday, December 2, 2013 at 2:12pm.

2y=4sqrt(x) , y=3, and 2y+3x=7

- calculus area between three curves -
**jaime**, Monday, December 2, 2013 at 2:22pmalso help with

Find c>0 such that the area of the region enclosed by the parabolas

y=x^2-c^2 and y=c^2-x^2 is 430.

- calculus area between three curves -
**Reiny**, Monday, December 2, 2013 at 2:32pm3 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+%28-3%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 right-angled 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**, Monday, December 2, 2013 at 2:39pmnice symmetry, should be easy to see that the x-intercepts 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

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