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
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 $\displaystyle \int_{1}^2 x-1\,dx + \int_{2}^{2\sqrt{2}} 2-\frac{x^2}{4}\,dx = \frac{1}{2}+\frac{2}{3}(4\sqrt{2}+5) = \frac{1}{6}(16\sqrt{2}-17)$.
I hope this helps you in some way.
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.
Well, as y goes from $\displaystyle 1$ to $\displaystyle 2$, x goes from $\displaystyle y$ to $\displaystyle 2\sqrt{y}$. (If $\displaystyle y=\frac{x^2}{4}$, then $\displaystyle x=\sqrt{4y}=2\sqrt{y}$).
How about this: $\displaystyle \int_{1}^{2} \int_{y}^{2\sqrt{y}} x\,dx\,dy$?
Integrating with respect to $\displaystyle x$ gives you $\displaystyle \frac{x^2}{2}$. Plugging in the bounds gives you $\displaystyle \frac{(2\sqrt{y})^2}{2}-\frac{y^2}{2} = 2y-\frac{y^2}{2}$.
Now the integral is $\displaystyle \int_{1}^{2} 2y-\frac{y^2}{2} \,dy$.
You can integrate this by normal means to get $\displaystyle y^2-\frac{y^3}{6}$. Plugging in the bounds gives you $\displaystyle 2^2-\frac{2^3}{6} - (1^2-\frac{1^3}{6}) = 4-\frac{8}{6} - 1+\frac{1}{6} = \frac{11}{6}$.
So $\displaystyle \frac{11}{6}$ is the answer you're looking for.