1. ## Triple integral, doubt

Hi there, I have this iterated integral $\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{2-y}\displaystyle\int_{0}^{4-y^2}dxdzdy$, and the thing is, does it define a solid? because I think that as it is given it doesn't, but I'm not sure. I think that the cylindric paraboloid never cuts the x axis, and thats why I think this integral doesn't define any solid.

2. The integral has sense in the region:

$T\equiv{}\begin{Bmatrix}0\leq{y}\leq{2}\\ 0\leq{z}\leq{2-y} \\ 0\leq{x}\leq{4-y^2}\end{matrix}$

Besides:

$\displaystyle\iiint_{T}dV=\displaystyle\int_{0}^{2 }\displaystyle\int_{0}^{2-y}\displaystyle\int_{0}^{4-y^2}dxdzdy=\displaystyle\int_{0}^{2}dy\displaystyl e\int_{0}^{2-y}dz\displaystyle\int_{0}^{4-y^2}dx$

Regards.

3. Originally Posted by Ulysses
Hi there, I have this iterated integral $\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{2-y}\displaystyle\int_{0}^{4-y^2}dxdzdy$, and the thing is, does it define a solid? because I think that as it is given it doesn't, but I'm not sure. I think that the cylindric paraboloid never cuts the x axis, and thats why I think this integral doesn't define any solid.
What? $x= 4- y^2$ cuts the x-axis when x= 4.

4. I'm starting to figure it out now. As the concavity is pointing at the x axis I thought it wouldn't cut ever the other integrands, but now I think I'm seeing it, the base would be at the yz plane I think. I'll try again.

Thank you both.