The surface area of a spherical part of a bow | vedclass

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Question

(D) Let the radius of the spherical bowl be $r \, cm$.The surface area of the hemispherical part (excluding the cover) is given by $2 \pi r^2 = 616 \,

Vedclass · Accepted Answer

$2032.69$

Vedclass · Answer

(D) Let the radius of the spherical bowl be $r \, cm$.The surface area of the hemispherical part (excluding the cover) is given by $2 \pi r^2 = 616 \, cm^2$.$r^2 = \frac{616 	imes 7}{2 	imes 22} = \frac{308 	imes 7}{22} = 14 	imes 7 = 98 \, cm^2$.$r = \sqrt{98} = 7\sqrt{2} \, cm$.The volume of a hemispherical bowl is given by $V = \frac{2}{3} \pi r^3$.$V = \frac{2}{3} 	imes \frac{22}{7} 	imes (7\sqrt{2})^3$.$V = \frac{2}{3} 	imes \frac{22}{7} 	imes 343 	imes 2\sqrt{2}$.$V = \frac{44}{21} 	imes 686\sqrt{2} = \frac{44 	imes 98\sqrt{2}}{3} = \frac{4312\sqrt{2}}{3}$.Using $\sqrt{2} \approx 1.414$,$V \approx \frac{4312 	imes 1.414}{3} \approx 2032.69 \, cm^3$.

The surface area of a spherical part of a bowl with a flat circular detachable cover,excluding the cover,is $616 \, cm^2$. The area of the cover is $38.5 \, cm^2$. What is the volume of the bowl? (in $cm^3$)

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