Calculus: Area
posted by jasmineT on .
Can someone look at my work and see what i did wrong. I did this 100 times but i keep getting it wrong
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=3x^(1/2) , y=4 and 2y+1x=4
y=3(x^(1/2))/2, y=4 , y=2(1/2)x
Intersection points = 4,1,64/9
Integral from 4 to 1
4x2x(x^2)/4
Integral from 1 to 64/9
4x3(x^(5/2))/5
please help i've been working on this for hours

go to wolframalpha . com and enter
graph x=4y^2/9,x+2y=4,y=4,y=3/2,x=1
to see the curves involved.
The area of interest is a little vshaped chunk from
(4,4) down to (1,3/2) and back up to (64/9,4)
If you integrate on x, then the integrands use the limits you have, but
the height in each case is 4y, for the appropriate f(x)
4  (4x)/2 from 4 to 1 = 25/4
4  3/2 sqrt(x) from 1 to 64/9 = 175/27
match anything you got? 
Thank you soooo much steve i finally got it right!!!! :D