Consider the solid S described below. The base of S is the triangular region with vertices (0, 0), (2, 0), and (0, 2). Cross-sections perpendicular to the x-axis are squares. Find the volume V of this solid.

To answer that question, we need to get the equation of the hypothenuse (L) of the right triangle.

The solution is:

V = 4/3 cubic units

As the cross-sections perpendicular (pieces) to the x-axis are squares, the volume (V) of the solid,  is the product of the area of the base ( the square ) times dx. We will integrate from 0 to the point ( 2 , 0 ).

Then

The straight line passing through P ( 2, 0 ) and Q ( 0 , 2) is:

y = m×x + b

we need to find m and b

m = ( 2 - 0 ) / ( 0 - 2 )     m = - 1

P  ( 2, 0 )      ⇒    y = m×x + b  ⇒   0 = -1× (2) + b       ⇒  b = 2

The equation of the hypothenuse is:

y = - x  + 2           ⇒   x = 2 - y

And  this is the side of each square, then

dV = A × dx    ⇒  ∫dV = ∫ A×dx

V = ∫ A×dx    

A is the area of the square of side:  y = 2 - x; then

V = ∫ ₀² ( 2 - x )² × dx

V = ∫ ₀² ( 4 + x² -4×x ) × dx

V = 4×x + (1/3)×x³ - 2×x² |₀²      V = 4×2 + (1/3)×(2)² - 4×(2) - 0

V = 4/3 cubic units

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is the region in the plane bounded by lines , , and or .

Each cross section, for a given value of in the interval [0, 2], has an area equal to , where is the length of the square section's base.

Then the volume of the solid is

step-by-step explanation:

the domain of a function is the set of input values, x , for which a function is defined. the domain is shown in the left oval in the picture below. the function provides an output value, f(x) , for each member of the domain.