Answer :
check the picture below.
so the lunchbox is really a semi-circle with a radius of 3, sitting on top of a square of 6x6, with a lenght of 10.
the volume of the lunchbox then will be the area of the semi-circle and square times the length.
[tex]\bf \left[ \stackrel{\textit{semi-circle's area}}{\frac{\pi 3^2}{2}}+\stackrel{\textit{square's area}}{(6\cdot 6)} \right]\stackrel{length}{10}\implies \left( \frac{9\pi }{2}+36\right)(10)[/tex]
so the lunchbox is really a semi-circle with a radius of 3, sitting on top of a square of 6x6, with a lenght of 10.
the volume of the lunchbox then will be the area of the semi-circle and square times the length.
[tex]\bf \left[ \stackrel{\textit{semi-circle's area}}{\frac{\pi 3^2}{2}}+\stackrel{\textit{square's area}}{(6\cdot 6)} \right]\stackrel{length}{10}\implies \left( \frac{9\pi }{2}+36\right)(10)[/tex]
