For a given [tex]x[/tex] in the interval [0, 1], the corresponding cross-section is a semicircle with diameter [tex]d=1-x^3[/tex], so the area of the cross-section would be [tex]\frac{\pi d^2}8[/tex].
The volume of the solid is then
[tex]\displaystyle\frac\pi8\int_0^1(1-x^3)^2\,\mathrm dx=\boxed{\frac{9\pi}{112}}[/tex]
The volume of a solid is the amount of space in it.
The volume of the solid is: [tex]\mathbf{ \frac{9\pi}{112} }[/tex]
The given parameters are:
[tex]\mathbf{y =x^3,\ \ y=1}[/tex]
From the question, we understand that the cross-sections are semicircles.
This means that, the diameter (d) is:
[tex]\mathbf{d = y_2 -y_1}[/tex]
So, we have:
[tex]\mathbf{d = 1 -x^3}[/tex]
The volume (V) of the solid is then calculated as:
[tex]\mathbf{V = \frac{\pi}{8} \int\limits^a_b {d^2} \, dx }[/tex]
Substitute [tex]\mathbf{d = 1 -x^3}[/tex]
[tex]\mathbf{V = \frac{\pi}{8} \int\limits^1_0 {(1 - x^3)^2} \, dx }[/tex]
Expand
[tex]\mathbf{V = \frac{\pi}{8} \int\limits^1_0 {(1 - 2x^3 + x^6)} \, dx }[/tex]
Integrate, with respect to x
[tex]\mathbf{V = \frac{\pi}{8} {(x - \frac{x^4}{2} + \frac{x^7}{7})} \, |\limits^1_0 }[/tex]
Substitute 1 and 0, for x
[tex]\mathbf{V = \frac{\pi}{8} [(1 - \frac{1^4}{2} + \frac{1^7}{7}) - (0 - \frac{0^4}{2} + \frac{0^7}{7})]}[/tex]
[tex]\mathbf{V = \frac{\pi}{8} [(1 - \frac{1}{2} + \frac{1}{7}) ]}[/tex]
Take LCM
[tex]\mathbf{V = \frac{\pi}{8} [\frac{14 - 7 + 2}{14} ]}[/tex]
[tex]\mathbf{V = \frac{\pi}{8} [\frac{9}{14} ]}[/tex]
Simplify the fraction
[tex]\mathbf{V = \frac{9\pi}{112} }[/tex]
Hence, the volume of the solid is: [tex]\mathbf{ \frac{9\pi}{112} }[/tex]
Read more about volumes at:
https://brainly.com/question/7251133
