[tex]S[/tex] is the region in the [tex]x,y[/tex] plane bounded by lines [tex]x=0[/tex], [tex]y=0[/tex], and [tex]x+y=2[/tex] or [tex]y=2-x[/tex].
Each cross section, for a given value of [tex]x[/tex] in the interval [0, 2], has an area equal to [tex]s^2[/tex], where [tex]s=(2-x)-0=2-x[/tex] is the length of the square section's base.
Then the volume of the solid is
[tex]\displaystyle\int_0^2(2-x)^2\,\mathrm dx=\boxed{\frac83}[/tex]
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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