Vector Approach: Parallelepiped

• 1). Denote three adjacent sides meeting at one of the parallelepiped's corners as the three vectors "a", "b" and "c". Denote their lengths "A", "B" and "C". Denote the angle between "a" and "b" as "α" (alpha), the angle between "b" and "c" as "β" (beta) and the angle between "c" and "a" as "γ" (gamma).
  
  
• 2). Orient the parallelepiped (if only in your mind) so the face bounded by "a" and "b" is on the bottom. Using standard vector arithmetic, take the cross product of "a" and "b". The resulting cross product is the vector perpendicular to the face that vectors the "a" and "b" border. The length of the resulting vector equals the area of the bottom face. The reason is because the magnitude of the cross product equals "AB sin α", by the definition of the cross product. Since the shape of the parallelepiped is the same from the bottom face all the way up to the top, you only have left to multiply the height by the area of the bottom face.
  
  
• 3). Identify the height of the parallelepiped and multiply it by the area of the base. The result is the volume of the parallelepiped. In other words, the formula for the volume of a parallelepiped is the base area times the height. This is the same as taking the dot product of "c" and the cross product of "a" and "b". This is true because the dot product is, by definition, the length of "c" times the length of the cross product of "a" and "b" times the cosine of "θ". Here, "θ" is the angle between "c" and the vector perpendicular to the bottom face. In other words, "c cos θ" is the height of the parallelepiped. So the volume is "c*(axb)", if "*" stands for the dot product. If you don't know the height, but do know lengths "A", "B" and "C", and the angles "α", "β" and "γ", then proceed to the next section.
  
  

Trigonometric Approach: Parallelepiped

• 1). Solve for the unknown angle "θ" when you don't have the bounding sides in vector form by first recalling the trigonometric identity cx(axb) = a(c*b)-b(c*a).
  
  
• 2). Write the magnitude of the left-hand side of the equation as (ABsin α) C sin θ. Now one side of the equation can be written in terms of "θ". So you can eventually solve for "cos θ".
  
  
• 3). Take the dot product of the vector a(c*b)-b(c*a) with itself. This comes out to be (aCB cos β - bCA cos γ)*(aCB cos β - bCA cos γ), or (ABC)^2 [cos(2)β - 2 cos β cos γ cos α + cos(2)γ] (Recall that a*a=A^2 and a*b = AB cos α.) Here, cos(2)β means (cos β)^2. This result equals the square of the magnitude of the right-hand side of the equation in Step 1.
  
  
• 4). Square the result of step 2 to equate it with step 3. Eliminate the (ABC)^2 first for convenience. Therefore, sin(2)α sin(2)θ = cos(2)β - 2 cos β cos γ cos α + cos(2)γ.
  
  
• 5). Write "sin θ" in terms of "cos θ", since ABC sin α cos θ is the volume you want to solve for. Thus, sin(2)α[1-cos(2)θ] = cos(2)β - 2 cos β cos γ cos α + cos(2)γ.
  
  
• 6). Solve for "sin α cos θ" as follows:
  sin(2)α cos(2)θ = sin(2)α - cos(2)β + 2 cos β cos γ cos α - cos(2)γ = 1 - cos(2)α - cos(2)β + 2 cos β cos γ cos α - cos(2)γ.
  Therefore, the final formula for the volume of a parallelepiped is:
  ABC√[1 - cos(2)α - cos(2)β - cos(2)γ + 2 cos β cos γ cos α].
  
  

Calculus Approach: Truncated Square Pyramid

• 1). Determine if calculus would be a useful approach by deciding if the figure can be cut into thin wafers, each with the same shape. For example, a square pyramid with the pointed top cut off--leaving a surface parallel to the bottom face--is a stack of thin squares. The shape is constant, though the size varies.
  
  
• 2). Denote the base width of the truncated pyramid "B" and top width "T". Denote the height "H". Then the integral of the figure's volume is ∫x^2 dy, where "x" is the varying width of the squares, "dy" is the differential height of the squares and the relation between the two variables is y = H - H (x-T)/(B-T). You can see this because the sides of a pyramid are linear, so a linear equation applies. Furthermore, when "x" is "B", "y" must be "0". When "x" is "T", "y" must be "H".
  
  
• 3). Write the integral ∫x^2 dy in terms of one variable:
  dy equals -dx H/(B-T).
  The negative will come out in the wash upon integrating from x=B to x=T, which is smaller.
  
  
• 4). Take the integral. Recall that integrals are products of the integrand times the differential width. Taking the integral of a polynomial like x^2 is a simple matter of adding 1 to the exponent, then dividing by the new exponent. Next, plug in the two endpoints, "B" and "T", and take the difference between the two of them. In other words, the result is [-H/(B-T)][(T)^3/3 - (B)^3/3], or H/(B-T)[(B)^3/3 - (T)^3/3].