The base of a certain solid is the triangle with vertices at (-8,4), (4,4), and the origin. Cross-sections perpendicular to the y-axis are squares. How do you find the volume of the solid?

Question

Isabella Ackerman · Accepted Answer

Please see below.

Here is a picture (graph) of the base (in blue) with a thin slice taken perpendicular to the #y#-axis (in black).

The thinkness of this representative slice is #dy# so we need the sides of the square to be expressed in terms of #y# (not #x#).

The line on the right contains #(0,0)# and #(4,4)#. Its equation is #y = x# or #x = y#

The line on the left contains #(0,0)# and #(-8,4)#. Its equation is #y = -1/2x# or #x = -2y#

The side of the square built on the representative slice is #s = x_"greater"-x_"lesser" = x_"on the right"-x_"on the left"#.

So, #s = y - (-2y) = 3y#.

The volume of the representative slice is

#s^2 * "thickness" = (3y)^2 dy#.

The values of #y# vary from #0# to #4#, so the volume of the solid is

#V = int_0^4 (3y)^2 dy = {:3y^3]_0^4 = 192# (cubic units)

Axel Alvarez · Answer

undefined To find the volume of the solid, we need to determine the area of the base and then integrate the area to find the volume.

The base of the solid is a triangle with vertices at (-8,4), (4,4), and the origin. This triangle has a base of 12 units (from (-8,4) to (4,4)) and a height of 4 units (from the base to the origin along the y-axis). Therefore, the area of the base triangle is:

Area = 1/2 * base * height = 1/2 * 12 * 4 = 24 square units.

Since the cross-sections perpendicular to the y-axis are squares, the area of each square cross-section will be equal to the area of the base triangle.

To find the volume of the solid, we integrate the area of the base over the range of y-values for which the solid exists. In this case, the solid exists from y = 0 to y = 4 (the height of the base triangle).

Therefore, the volume of the solid is:

Volume = ∫[0,4] 24 dy = 24y ∣[0,4] = 24(4) - 24(0) = 96 cubic units.