\(\displaystyle Hint:\)
\(\displaystyle A \ wedge \ is \ cut \ from \ an \ 18-inch \ diameter \ by \ first \ making \ a \ horizontal \ cut \ 9 \ inches\)
\(\displaystyle \ deep, \ then \ a \ slanted \ cut \ starting \ at \ a \ point \ 9 \ inches \ above \ the \ horizontal \ cut. \ Set\)
\(\displaystyle \ up \ an \ integral \ for \ the \ volume \ of \ the \ wedge.\)
\(\displaystyle Now, \ if \ the \ above \ was \ the \ case, \ we \ proceed \ accordingly.\)
\(\displaystyle The \ cross \ sections \ are \ isosceles \ right \ triangles \ for \ which \ the \ base \ and \ height \ are \ equal.\)
\(\displaystyle Hence, \ A(x) \ = \ \bigg(\frac{1}{2}\bigg)bh \ = \ \bigg(\frac{1}{2}\bigg)(\sqrt{81-x^{2}})(\sqrt{81-x^{2}}) \ = \ \bigg(\frac{1}{2}\bigg)(81-x^{2})\)
\(\displaystyle Note: \ x^{2} \ + \ y^{2} \ = \ 81, \ y \ = \ \sqrt{81-x^{2}}\)
\(\displaystyle Ergo, \ V \ = \ \bigg(\frac{1}{2}\bigg)\int_{-9}^{9}(81-x^{2})dx \ = \ \int_{0}^{9}(81-x^{2})dx \ = \ 486 \ cu. \ in.\)
\(\displaystyle However, \ in \ your \ problem, \ the \ wedge \ is \ cut \ 3 \ inches \ short \ of \ the \ diameter.\)
\(\displaystyle Therefore, \ V \ = \ \int_{0}^{6\sqrt2}[\sqrt{81-x^{2}}-3]^{2}dx \ = \ about \ 185 \ cu. \ in.\)
