# Thread: Double integral region bound by curves

1. ## Double integral region bound by curves

I need to find the volume of a function x dA where S is bound by the curves y = x^2/4, y=x, y=1,and y=2 with x greater or equal to 0

I think x goes from 0 to 2 would y be between x and x^2/4???? Frostking

2. I've attached a picture of the graphs of these. I'd like to help you, but I'm not sure what you're asking. Is this a rotation problem? The area bounded by the curves is $\int_{1}^2 x-1\,dx + \int_{2}^{2\sqrt{2}} 2-\frac{x^2}{4}\,dx = \frac{1}{2}+\frac{2}{3}(4\sqrt{2}+5) = \frac{1}{6}(16\sqrt{2}-17)$.

I hope this helps you in some way.

3. ## Clarification of double integral problem

I am sorry I was not clear. The exact question is : FInd the double integral over the region S of x dA where S is the region bounded by the curves y = x^2/4, y = x, y =1 and y = 2 with x greater than or equal to 0.

4. Well, as y goes from $1$ to $2$, x goes from $y$ to $2\sqrt{y}$. (If $y=\frac{x^2}{4}$, then $x=\sqrt{4y}=2\sqrt{y}$).

How about this: $\int_{1}^{2} \int_{y}^{2\sqrt{y}} x\,dx\,dy$?

Integrating with respect to $x$ gives you $\frac{x^2}{2}$. Plugging in the bounds gives you $\frac{(2\sqrt{y})^2}{2}-\frac{y^2}{2} = 2y-\frac{y^2}{2}$.

Now the integral is $\int_{1}^{2} 2y-\frac{y^2}{2} \,dy$.

You can integrate this by normal means to get $y^2-\frac{y^3}{6}$. Plugging in the bounds gives you $2^2-\frac{2^3}{6} - (1^2-\frac{1^3}{6}) = 4-\frac{8}{6} - 1+\frac{1}{6} = \frac{11}{6}$.

So $\frac{11}{6}$ is the answer you're looking for.