# Thread: Double integrals to find the volume of a solid

1. ## Double integrals to find the volume of a solid

Example 3. Find the volume of the solid enclosed by surface y^2 = x and by planes z = 0 and x + z = 1.

I am stuck on what to put for limits, I keep getting -infinity for my answer.

-x+1 dxdy I get 0 to -infinity for the first integral and -sqrt y to + sqrt y for the second integral.

2. Solve x+z=1 for z and get z=1-x.

$\int_{-1}^{1}\int_{y^{2}}^{1}(1-x)dxdy$

3. $y^2= x$ is a "parabolic cylinder" opening in the positive x direction. It is bounded below by z= 0 and above by z= 1- x. Fortunately those two intersect in the line x= 1, bounding the cylinder on the right. In the x,y plane, the figure is bounded by $y^2= x$ on the left, top, and bottom, and by x= 1 on the right.

You could take x from 0 to 1 and, for each x, $y= -\sqrt{x}$ to $y= \sqrt{x}$. For each x,y, z goes from 0 up to 1- x so the integrand is 1-x-0= 1- x.

The volume is given by $\int_{x= 0}^1\int_{y= -\sqrt{x}}^{\sqrt{x}} (1- x) dy dx$.

But, of course, it is easier to reverse the order of integration. The lowest y value on the entire figure is -1 at (1,-1) and the highest is 1 at (1,1), both where the parabola intersects the line x= 1. For each y, then x ranges from the parabola $x= y^2$ to the line x= 1.

The volume is given by $\int_{y=-1}^1\int_{x= y^2}^1 (1- x) dx dy$.