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

Answer:

$V = \frac{4y^3}{3} -12y^2 +36 y \Big|_0^3$

$V= \frac{4(3)^3}{3} -12(3)^2 +36 (3) -0 = 36-108+108= 36$

Step-by-step 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:

$V = \frac{4y^3}{3} -12y^2 +36 y \Big|_0^3$

$V= \frac{4(3)^3}{3} -12(3)^2 +36 (3) -0 = 36-108+108= 36$