# Thread: 2 quick questions on double integrals

1. ## 2 quick questions on double integrals

I'm having a bit of trouble finding the the bounds for the following problems.

1) $z=x^2+3y^2$ and the planes $x = 0, \ y=1, \ y=x, \ z=0$

since z will be a paraboloid and bounded by those constraints I'm thinking I would have $\int_{x=0}^{x=1} \int _{y=x}^{y=1} x^2+3y^2 \ dy \ dx$

2) $z= x^2 , \ y=x^2$ and the planes $z=0, \ y=4$

so I get the graph that's in the attachment, now if I take those constraints into account then I would simple have something above the xy plane and to the right of the xz plane, therefore I would think that my integral would be:

$\int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} x^2 \ dy \ dx$

2. Originally Posted by lllll
I'm having a bit of trouble finding the the bounds for the following problems.

1) $z=x^2+3y^2$ and the planes $x = 0, \ y=1, \ y=x, \ z=0$

since z will be a paraboloid and bounded by those constraints I'm thinking I would have $\int_{x=0}^{x=1} \int _{y=x}^{y=1} x^2+3y^2 \ dy \ dx$

2) $z= x^2 , \ y=x^2$ and the planes $z=0, \ y=4$

so I get the graph that's in the attachment, now if I take those constraints into account then I would simple have something above the xy plane and to the right of the xz plane, therefore I would think that my integral would be:

$\int_{x=-2}^{x=2} \int_{y=x^2}^{y=4} x^2 \ dy \ dx$
If you're wishing to find the volume under your given z's and above the $xy$ plane, then your integrals are correct!