Question

The base of a solid is the region in the first quadrant bounded by the line x=-2y+6 and the coordinate axes. What is the volume of the solid if every cross section perpendicular to the y-axis is a square. .

Answer 1

`V = (4y^3)/(3) -12y^2 +36 y \Big|_0^3`

`V= (4(3)^3)/(3) -12(3)^2 +36 (3) -0 = 36-108+108= 36`

Explanation:

For this case we have the following plot on the figure attached.

We know on this case that
`x = -2y+6` represent our radius for the square, since we need cross sections perpendicular to the y axis.

We find the intersection points like this:

`x= 0 , y=3`

`y=0, x=6`

Since we are assuming squares for the cross sections the area is given by
`A = r^2` where r = x= -2y+6.

Since our radius is on terms of x we can create a integral with limits on x in order to find the volume, And we can use the following integral in order to find the volume.

`V = \int_(0)^3 (-2y+6)^2 = \int_(0)^3 4y^2 -24y +36 dy`

On the left part we are using the following property from algebra:

`(a+b)^2 = a^2 +2ab +b^2`

`(-2y+6)^2 = (-2y)^2 +2*(-2y)(6) +(6)^2 = 4y^2 -24y+36`

And after do the integral we got this:

`V = (4y^3)/(3) -12y^2 +36 y \Big|_0^3`

`V= (4(3)^3)/(3) -12(3)^2 +36 (3) -0 = 36-108+108= 36`