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Since we know the intersecting of the sphere of radius and the plain is a circle of radius r.

Then the circle is

x^{2} + y^{2} = r^{2}

Clearly, the circle has ends x = -r to +r

Now, the volume of the solid that is generated by revolving the circle (i) about x-axis (i.e. y = 0) is

\(Volume = \pi \int_{x=-r}^{r} y^2dx = \pi \int_{-r}^{r} (r^2 – x^2)dx\)

\(= \pi \left [ r^2 \times – \frac{x^3}{3} \right ]^r_{-r}\)

\(= \pi \left[ \left ( r^3 – \frac{r^3}{3} \right ) – \left ( – r^3 + \frac{r^3}{3} \right ) \right ]\)

\(= \pi \left [ \frac{2r^3}{3} + \frac{2r^3}{3} \right ]\)

\( = \frac{4\pi r^3}{3}\)

Since the solid that is generated by revolving a circle (i) about x-axis, is a sphere.

Therefore, The volume of the sphere whose radius r is \(\frac{4\pi r^3}{3}\).

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