Compute the volume of a 3-d object with a trapezoidal 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

Are you sure this isn't a pentagonal base?

Answer 2

No one said you have to eat your chicken in one piece. And so it is here also.

With kudos to Doug C. I shall refer to his graphic, https://www.desmos.com/calculator/wtbxnhb5lp

The volume of the rectangular parallelepiped (x = 6 through x = 9) is simple arithmetic.

The volume of the triangular section is the attention getter.

The area of each square is (6 - x)², or 36 - 12x + x² for x = 0 to x = 6.

Integrate for the volume, multiply by 2, and add the volume of the rectangular parallelepiped.

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.