Question

Find the volume of material remaining in a hemisphere of radius 2 after a cylindrical hole ofradius 1 is drilled through the center of the hemisphere perpendicular to its base.

Answer 1

Answer:

Step-by-step explanation:

Inicially we have to separate the sphere into 3 parts:

The donut  +  1 ball cap (or a dome) +  the cylinder hole = the Hemisphere

vide picture

As it shows in the picture we have to find the donut.

The donut  = the Hemisphere - ( 1 dome +  the cylinder hole)

• The hemisphere = $\frac{4*pi*r^{3} }{6}$ (from the hemisphere formula)
• 1 dome* = $\frac{1}{6}*pi*(r-\sqrt{(r^{2} -a^{2}})*(3a^{2}+((r-\sqrt{(r^{2} -a^{2}})^{2} )$
• cylinder hole = $pi*a^{2}*h = pi*a^{2} *(2r - 2*(r-\sqrt{(r^{2} -a^{2}}))$ (h: cylinder height)

$D = \frac{4*pi*r^{3} }{6} - (\frac{1}{6}*pi*(r-\sqrt{(r^{2} -a^{2}})*(3a^{2}+((r-\sqrt{(r^{2} -a^{2}})^{2} )+ pi*a^{2}*(2r - 2*(r-\sqrt{(r^{2} -a^{2}})) }$

$D = \frac{4*pi*2^{3} }{6} - (\frac{1}{6}*pi*(2-\sqrt{(2^{2} -1^{2}})*(3*1^{2}+((2-\sqrt{(2^{2} -1^{2}})^{2} )+ pi*1^{2}*(2*2 - 2*(2-\sqrt{(2^{2} -1^{2}})) }$

By basic algebra we have

D = 5.304

*The dome in the 1st picture is represented by the color orange.Looking into the second picture, the dome formula derivatives from the volume of a dome of a hemisphere. By using the following linear system:

• triangle_side2 + a2 = r2
• h + triangle_side = r