Use a double integral to find the area of the region whose upperbound is the function y = (x+3)^1/2 the lower boundary is y = (x+3)/3

I got 4.5

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- Jun 2nd 2010, 09:01 AMewkimchiUse a double integral to find the area of the region whose upperbound is the function
Use a double integral to find the area of the region whose upperbound is the function y = (x+3)^1/2 the lower boundary is y = (x+3)/3

I got 4.5 - Jun 2nd 2010, 10:20 AMAmer
find the point of the intersection

$\displaystyle \sqrt{x+3}=\frac{x+3}{3} $

$\displaystyle 9(x+3) = x^2 + 6x+9 $

$\displaystyle x^2 -3x-18 = 0 $

$\displaystyle (x-6)(x+3) = 0 \Rightarrow x=6,x=-3 $

$\displaystyle \int_{6}^{-3} \int_{\frac{x+3}{3}}^{\sqrt{x+3}} dy dx $

I got 4.5 to