# Math Help - Volume of a solid with known cross sections

1. ## Volume of a solid with known cross sections

"Any cross sectional slice of a certain solid in a plane perpendicular to the x-axis is a square with side AB, with A lying on the curve $y^2 = 4x$ and B on the curve $x^2 = 4y$. Find the volume of the solid lying between the points of intersection of these two curves."

I'm not sure if I'm going in the right direction, but so far I've put the curves in terms of y, leaving me with $y = 2\sqrt{x}$ and $y = \frac{x^2}{4}$. After graphing, I also know that the limits of integration will be from 0 to 4 since the points of intersection are at (0, 0) and (4, 4). From here on, I'm completely lost.

Thanks

"Any cross sectional slice of a certain solid in a plane perpendicular to the x-axis is a square with side AB, with A lying on the curve $y^2 = 4x$ and B on the curve $x^2 = 4y$. Find the volume of the solid lying between the points of intersection of these two curves."

I'm not sure if I'm going in the right direction, but so far I've put the curves in terms of y, leaving me with $y = 2\sqrt{x}$ and $y = \frac{x^2}{4}$. After graphing, I also know that the limits of integration will be from 0 to 4 since the points of intersection are at (0, 0) and (4, 4). From here on, I'm completely lost.

Thanks
That is a good start and we have been told that our solid is a square.

Using your graph you will note that the side length of one side of the square is

$2\sqrt{x}-\frac{x^2}{4}$

Now to find the volume we will find the area of the square and multiply by dx to get a volume

$\int_{0}^{4} \left( 2\sqrt{x}-\frac{x^2}{4}\right)^2dx$