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# geometry

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A spherical soap bubble lands on a horizontal wet surface and forms a hemisphere of the same volume. Given the radius of the hemisphere is $3\sqrt[3]{2}$cm, find the radius of the original bubble.

#1
+4694
+1

volume of sphere  $$=\,\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3$$

volume of hemisphere  $$=\,\frac12\,*\,\frac43\,*\,\pi\,*\, (\text{radius of hemisphere})^3$$

Plug in  $$3\sqrt[3]2$$  for the radius of the hemisphere.

volume of hemisphere  $$=\,\frac12\,*\,\frac43\,*\,\pi \,*\,(3\sqrt[3]2)^3 \\~\\ =\,\frac23\,*\,\pi\,*\,3^3\,*\,\sqrt[3]2^3 \\~\\ =\,\frac23\,*\,\pi\,*\,27\,*\,2 \\~\\ =\,36\pi$$

The hemisphere has the same volume as the sphere, so....

volume of sphere  =  volume of hemisphere

$$\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3\,=\,36\pi$$

Divide both sides by  pi .

$$\frac43\,*\, (\text{radius of sphere})^3\,=\,36$$

Multiply both sides by  3/4  .

$$(\text{radius of sphere})^3\,=\,27$$

Take the cube root of both sides.

$$\text{radius of sphere}\,=\,3\,\text{ cm}$$

hectictar  Sep 24, 2017
Sort:

#1
+4694
+1

volume of sphere  $$=\,\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3$$

volume of hemisphere  $$=\,\frac12\,*\,\frac43\,*\,\pi\,*\, (\text{radius of hemisphere})^3$$

Plug in  $$3\sqrt[3]2$$  for the radius of the hemisphere.

volume of hemisphere  $$=\,\frac12\,*\,\frac43\,*\,\pi \,*\,(3\sqrt[3]2)^3 \\~\\ =\,\frac23\,*\,\pi\,*\,3^3\,*\,\sqrt[3]2^3 \\~\\ =\,\frac23\,*\,\pi\,*\,27\,*\,2 \\~\\ =\,36\pi$$

The hemisphere has the same volume as the sphere, so....

volume of sphere  =  volume of hemisphere

$$\frac43\,*\,\pi\,*\, (\text{radius of sphere})^3\,=\,36\pi$$

Divide both sides by  pi .

$$\frac43\,*\, (\text{radius of sphere})^3\,=\,36$$

Multiply both sides by  3/4  .

$$(\text{radius of sphere})^3\,=\,27$$

Take the cube root of both sides.

$$\text{radius of sphere}\,=\,3\,\text{ cm}$$

hectictar  Sep 24, 2017
#2
+76821
+1

Nice, hectictar.....!!!!

CPhill  Sep 24, 2017

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