Question

Find the volume v of the described solid s. The base of s is a circular disk with radius r. Parallel cross-sections perpendicular to the base are squares.

Answer 1

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

Let S be solid lying between z=a and z=b. If the cross-sectional area of S in the plane Px through x and perpendicular to the x-axis is A(x)

Volume of S = $lim_{n- > \infty}\Sigma_{i=1}^{n}$ A($x_{i}^{*}$) Δx = $\int\limits^b_a {A(x)} \, dx$

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 = $\int\limits^r_{-r}{A(x)} \, dx$

Substitute A(x) from (1)

V = $\int\limits^r_{-r}{4(r^2 - x^2)} \, dx$

⇒V = 4[r²x - x³/3 ] $|_{-r}^{r}$

⇒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

Integrals to find volume:

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