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Sagot :
The volume of the described solid is V = 16r³/3
Given,
A solid line S should be drawn between z=a and z=b. If S has a cross-sectional area of Px through x and is perpendicular to the x-axis, then A (x)
Volume of S = limₙ→∝ ∑i=₁ⁿ A(xi) Δx = [tex]\int\limits^b_a {A(x)} \, dx[/tex]
The equation of circle x² + y² = r² where r is the radius of the circle.
Equation of upper semicircle yu= √(r²-x²) and
Equation of lower semicircle yl = -√(r²-x²)
Area of cross-section A(x) = (yu² - yl²)
Substitute yu and yl values
A(x) = [√(r²-x²) - (-√(r²-x²) )]² = 4(r² -x²) (1)
The limit for volume v varies from -r to r
Thus Volume v = ₋[tex]\int\limits^r_r {A(x)} \, dx[/tex]
Substitute A(x) from (1)
V = ₋[tex]\int\limits^r_r {4(r^{2}-x^{2} ) } \, dx[/tex]
⇒V = 4[r²x - x³/3 ]
⇒V = 4(r³ - r³/3) - 4(-r³ + r³/3) = 4[2 (r³ - r³/3)] = 16r³/3
The volume of the described solid S where the base is a circular disk or radius r and parallel cross sections perpendicular to the base are squares is V = 16r³/3
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