Answer:
[tex] V = \frac{4y^3}{3} -12y^2 +36 y \Big|_0^3[/tex]
[tex] V= \frac{4(3)^3}{3} -12(3)^2 +36 (3) -0 = 36-108+108= 36[/tex]
Step-by-step explanation:
For this case we have the following plot on the figure attached.
We know on this case that [tex] x = -2y+6[/tex] represent our radius for the square, since we need cross sections perpendicular to the y axis.
We find the intersection points like this:
[tex] x= 0 , y=3[/tex]
[tex] y=0, x=6[/tex]
Since we are assuming squares for the cross sections the area is given by [tex] A = r^2[/tex] 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.
[tex] V = \int_{0}^3 (-2y+6)^2 = \int_{0}^3 4y^2 -24y +36 dy[/tex]
On the left part we are using the following property from algebra:
[tex] (a+b)^2 = a^2 +2ab +b^2[/tex]
[tex](-2y+6)^2 = (-2y)^2 +2*(-2y)(6) +(6)^2 = 4y^2 -24y+36[/tex]
And after do the integral we got this:
[tex] V = \frac{4y^3}{3} -12y^2 +36 y \Big|_0^3[/tex]
[tex] V= \frac{4(3)^3}{3} -12(3)^2 +36 (3) -0 = 36-108+108= 36[/tex]
