Volume of the solid bounded by $4x^2 + y^2 =4$ , $z=o$ and $z=x+5$ I need to find the volume of the solid bounded by the cylinder $4x^2 + y^2 =4$ ,  $z=o$ and the plane $z=x+5$. I know that $0 \le z \le x+5$ , $-2 \sqrt{1-x^2} \le y\le 2 \sqrt{1-x^2}$ and $-1 \le x\le 1$. This gives me $\int_{-1}^1 \int_{-2 \sqrt{1-x^2}} ^{2 \sqrt{1-x^2}} \int_0 ^{x+5} dzdydx$.  
I'm stuck here. I think I need to use cylindrical coordinates but I'm not sure.
Thanks.

The first integral gives you, $(x+5)$. 
The second gives you, $y(x+5)|_{y=-2\sqrt{1-x^2}}^{y=2\sqrt{1-x^2}} = 4\sqrt{1-x^2}(x+5)$. 
Now just take the integral of this with respect to $x$, remembering that $z$ is to be considered a constant. 
Solve: $4 \int_{-1}^{1} \sqrt{1-x^2}(x+5) dx$. Use $u=1-x^2$
Let me know if this doesn't work out for you :).
