Compute the volume of a 3-d object with a pentagonal base (the base's vertices are (0,0), (6,3), (6,-3), (9,3) & (9,-3)), given that perpendicular cross sections are squares.

Answer 1

The base is symmetric to the x-axis. So, we can find the volume of the portion of the solid with base in the first quadrant and then double the result. I am assuming that the cross sections are taken perpendicular to the x-axis.

Between x = 6 and x = 9, the portion of the solid that lies above the first quadrant is a cube with

side length 3. So, the volume of the solid between x = 6 and x = 9 is 2(3)(3)(3) = 54.

Between x = 0 and x = 6, the base of the solid that lies in the first quadrant is a right triangle. An equation of the hypotenuse is y = (1/2)x.

Volume of solid between x = 0 and x = 6 is 2∫_{(0 to 6)} [(1/2)x]²dx = 36.

Total volume of solid = 36 + 54 = 90.

Answer 2

base area = sum of a rectangle and triangle

square is 6 by 3 = 18

triangle is 6 by 6 with area = 36/2 = 18

base area = 18+18= 36

helps to graph the 5 points construct the rectangle & triangle

square perpendicular cross sections seems to mean height of the 3D object is same as length of the base =9

9 x 36 = 324 units^3 = volume

or maybe square cross sections mean height of 3 over the rectangle, then vary over the triangle from 0 to 3

then

volume of the object over the rectangle is 3 x18= 54

and volume over the triangle is of a pyramid on its side with base = 3 by 3, height =6

v=1/3 of base area time height = (3x3)x6)/3 = 18

total volume = 54+18 = 72 units^3

while the problem seems a little ambiguous I'm inclined to go with 72

Answer 3

To compute the volume of the 3D object with a trapezoidal base, where perpendicular cross sections are squares, we can integrate the area of each square cross section along the length of the trapezoidal base.

The area of a square is given by \(A_{\text{square}} = s^2\), where \(s\) is the side length of the square.

The length of the trapezoidal base is from \(x = 0\) to \(x = 9\), and at each \(x\), the side length of the square is determined by the height of the trapezoid.

Let's denote the height of the trapezoid at \(x\) as \(h(x)\), which can be determined from the trapezoidal base's vertices.

The formula for the area of the square cross section at a given \(x\) is then \(A_{\text{square}}(x) = [h(x)]^2\).

The volume \(V\) of the 3D object is given by the integral of the area of the squares along the length of the trapezoidal base:

\[V = \int_{0}^{9} [h(x)]^2 \,dx\]

Now, let's determine the expression for \(h(x)\) based on the given vertices of the trapezoidal base.

The top of the trapezoid is formed by the points (6, 3) and (9, 3), and the bottom by (0, 0), (6, -3), and (9, -3). The height at each \(x\) is the difference in the \(y\)-coordinates of the corresponding top and bottom points.

\[h(x) = 3 - (-3) = 6\]

Now, we can set up and solve the integral:

\[V = \int_{0}^{9} [6]^2 \,dx\]

\[V = \int_{0}^{9} 36 \,dx\]

\[V = 36 \int_{0}^{9} dx\]

\[V = 36[x]_{0}^{9}\]

\[V = 36 \cdot (9 - 0)\]

\[V = 36 \cdot 9\]

\[V = 324\]

Therefore, the volume of the 3D object is 324 cubic units.