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(B) The equation of the circle representing the cross-section of the sphere is $x^2 + y^2 = 5^2 = 25$.The curved surface area $S$ of a solid of revolu

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$10\pi \, cm^2$

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(B) The equation of the circle representing the cross-section of the sphere is $x^2 + y^2 = 5^2 = 25$.The curved surface area $S$ of a solid of revolution generated by rotating the arc of the circle around the $x$-axis is given by $S = 2\pi \int_{x_1}^{x_2} y \, ds$,where $ds = \sqrt{1 + (\frac{dy}{dx})^2} \, dx$.From $x^2 + y^2 = 25$,we have $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$.Then $ds = \sqrt{1 + (-\frac{x}{y})^2} \, dx = \sqrt{\frac{y^2 + x^2}{y^2}} \, dx = \sqrt{\frac{25}{y^2}} \, dx = \frac{5}{y} \, dx$.The limits of integration are from $x_1 = 2 \, cm$ to $x_2 = 2 + 1 = 3 \, cm$.Thus,$S = 2\pi \int_{2}^{3} y \cdot \frac{5}{y} \, dx = 2\pi \int_{2}^{3} 5 \, dx$.$S = 2\pi [5x]_{2}^{3} = 2\pi (15 - 10) = 10\pi \, cm^2$.

$A$ frustum of a sphere is made by cutting a sphere with two parallel planes. If the radius of the sphere is $5 \, cm$ and the distance between the planes is $1 \, cm$,what is the curved surface area of the frustum when the distance of the first plane from the center of the sphere is $2 \, cm$?

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