Question

Find the volume of the solid whose base is the region in the first quadrant bounded by y=x3, y=1, and the y-axis and whose cross-sections perpendicular to the x axis are semicircles.

Answer 1

For a given $x$ in the interval [0, 1], the corresponding cross-section is a semicircle with diameter $d=1-x^3$, so the area of the cross-section would be $\frac{\pi d^2}8$.

The volume of the solid is then

$\displaystyle\frac\pi8\int_0^1(1-x^3)^2\,\mathrm dx=\boxed{\frac{9\pi}{112}}$

Answer 2

The volume of a solid is the amount of space in it.

The volume of the solid is: $\mathbf{ \frac{9\pi}{112} }$

The given parameters are:

$\mathbf{y =x^3,\ \ y=1}$

From the question, we understand that the cross-sections are semicircles.

This means that, the diameter (d) is:

$\mathbf{d = y_2 -y_1}$

So, we have:

$\mathbf{d = 1 -x^3}$

The volume (V) of the solid is then calculated as:

$\mathbf{V = \frac{\pi}{8} \int\limits^a_b {d^2} \, dx }$

Substitute $\mathbf{d = 1 -x^3}$

$\mathbf{V = \frac{\pi}{8} \int\limits^1_0 {(1 - x^3)^2} \, dx }$

Expand

$\mathbf{V = \frac{\pi}{8} \int\limits^1_0 {(1 - 2x^3 + x^6)} \, dx }$

Integrate, with respect to x

$\mathbf{V = \frac{\pi}{8} {(x - \frac{x^4}{2} + \frac{x^7}{7})} \, |\limits^1_0 }$

Substitute 1 and 0, for x

$\mathbf{V = \frac{\pi}{8} [(1 - \frac{1^4}{2} + \frac{1^7}{7}) - (0 - \frac{0^4}{2} + \frac{0^7}{7})]}$

$\mathbf{V = \frac{\pi}{8} [(1 - \frac{1}{2} + \frac{1}{7}) ]}$

Take LCM

$\mathbf{V = \frac{\pi}{8} [\frac{14 - 7 + 2}{14} ]}$

$\mathbf{V = \frac{\pi}{8} [\frac{9}{14} ]}$

Simplify the fraction

$\mathbf{V = \frac{9\pi}{112} }$

Hence, the volume of the solid is: $\mathbf{ \frac{9\pi}{112} }$

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