Answer:
The volume of the figure is [tex]\frac{l^{3}}{3}[\frac{1}{2}\pi-1]\ units^{3}[/tex]
Step-by-step explanation:
we know that
The volume of the figure is equal to the volume of the cone minus the volume of the square pyramid
step 1
Find the volume of the cone
The volume of the cone is equal to
[tex]V=\frac{1}{3}\pi r^{2} h[/tex]
we have
[tex]r=l\frac{\sqrt{2}}{2}\ units[/tex]
[tex]h=l\ units[/tex]
substitute
[tex]V=\frac{1}{3}\pi (l\frac{\sqrt{2}}{2})^{2}l[/tex]
[tex]V=\frac{1}{3}\pi (\frac{l^{3}}{2} )[/tex]
[tex]V=\frac{1}{6}\pi (l^{3})\ units^{3}[/tex]
step 2
Find the volume of the square pyramid
[tex]V=\frac{1}{3}Bh[/tex]
we have
[tex]B=l^{2}\ units^{2}[/tex]
[tex]h=l\ units[/tex]
substitute
[tex]V=\frac{1}{3}(l^{2})l[/tex]
[tex]V=\frac{1}{3}(l^{3})\ units^{3}[/tex]
step 3
Find the difference
[tex]\frac{1}{6}\pi (l^{3})-\frac{1}{3}(l^{3})=\frac{l^{3}}{3}[\frac{1}{2}\pi-1]\ units^{3}[/tex]
