Does Iterated Integral Define a Solid?

[Telemachus]

Homework Statement

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 that's why I think this integral doesn't define any solid.

 

[Mark44]

Homework Statement

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 that's why I think this integral doesn't define any solid.

If you sketch a graph of the region over which integration takes place, you'll see that it is a solid. The solid is in the first octant (i.e., bounded by the x-y plane, x-z plane, and y-z plane), and is bounded by the graph of the cylinder x = 4 - y² and the plane z = 2 - y.

 

[Telemachus]

Thank you very much Mark ;)