# Thread: volume of a region bounded by two surfaces

1. ## volume of a region bounded by two surfaces

I have this problem

find the voulme of the solid bounded above by surface z=50-4(y^2)
and below by z=9(y^2)+14

I only know how to calculate voules of solid when the side parameters are given(eg. bounded on sides by cylinder x^2+y^2=1)
but i dont know how to apply the double integral for solid bounded by two surfaces.

I would be grateful if someone can answer how to do it(or point to some website where it is clearly shown)

2. Originally Posted by Dili
I have this problem

find the voulme of the solid bounded above by surface z=50-4(y^2)
and below by z=9(y^2)+14

I only know how to calculate voules of solid when the side parameters are given(eg. bounded on sides by cylinder x^2+y^2=1)
but i dont know how to apply the double integral for solid bounded by two surfaces.

I would be grateful if someone can answer how to do it(or point to some website where it is clearly shown)

$V = \int \int_{R_{xy}} (50 - 4y^2) - (9y^2 + 14) \, dx \, dy \,$ where $\, R_{xy}\,$ is the region of the xy-plane defined by the infinite rectangular strip $-\frac{6}{\sqrt{13}} \leq y \leq \frac{6}{\sqrt{13}}$ and $-\infty < x < \infty$.
Note: the bounding values of y are found by solving $50 - 4y^2 = 9y^2 + 14$.