# Math Help - Use a double integral to find the area of the region whose upperbound is the function

1. ## Use 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

2. Originally Posted by ewkimchi
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
find the point of the intersection

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

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

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

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

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

I got 4.5 to