cylindrical wedge cut

[nhvt]

I have this problem which I have tried for 3 days and still get stuck. I just don't know how to visualize the problem. Can anyone help me to set it up?
A wedge is cut from an 18-inch diameter try by first making a horizontal cut 6 inches deep, then a slanted cut starting at a point 6 inches above the horizontal cut. Set up an integral for the volume of the wedge.

 

[tutor_joel]

First of all do you want to work this in Cartesian of cylindrical coordinates? Assuming Cartesian coordinates, this wedge is bounded by the xy plane (z = 0) and a plane at 45 degrees passing through the line y = 3. The base can be drawn using the point (0, 9, 0) and sketching the radius of 9 terminating at the line y =3. The second plane passing through y = 3 can be drawn using the point (0, 9, 6) and sketching in the same manner. Hopefully you can draw this out and see what it looks like since your main problem was visualizing this.

 

[nhvt]

I haven't studied about Cartesian of cylindrical coordinates, do you have any alternative ways to do this without using that?

 

[BigGlenntheHeavy]

$\displaystyle Hint:$

$\displaystyle A \ wedge \ is \ cut \ from \ an \ 18-inch \ diameter \ by \ first \ making \ a \ horizontal \ cut \ 9 \ inches$

$\displaystyle \ deep, \ then \ a \ slanted \ cut \ starting \ at \ a \ point \ 9 \ inches \ above \ the \ horizontal \ cut. \ Set$

$\displaystyle \ up \ an \ integral \ for \ the \ volume \ of \ the \ wedge.$

$\displaystyle Now, \ if \ the \ above \ was \ the \ case, \ we \ proceed \ accordingly.$

$\displaystyle The \ cross \ sections \ are \ isosceles \ right \ triangles \ for \ which \ the \ base \ and \ height \ are \ equal.$

$\displaystyle Hence, \ A(x) \ = \ \bigg(\frac{1}{2}\bigg)bh \ = \ \bigg(\frac{1}{2}\bigg)(\sqrt{81-x^{2}})(\sqrt{81-x^{2}}) \ = \ \bigg(\frac{1}{2}\bigg)(81-x^{2})$

$\displaystyle Note: \ x^{2} \ + \ y^{2} \ = \ 81, \ y \ = \ \sqrt{81-x^{2}}$

$\displaystyle Ergo, \ V \ = \ \bigg(\frac{1}{2}\bigg)\int_{-9}^{9}(81-x^{2})dx \ = \ \int_{0}^{9}(81-x^{2})dx \ = \ 486 \ cu. \ in.$

$\displaystyle However, \ in \ your \ problem, \ the \ wedge \ is \ cut \ 3 \ inches \ short \ of \ the \ diameter.$

$\displaystyle Therefore, \ V \ = \ \int_{0}^{6\sqrt2}[\sqrt{81-x^{2}}-3]^{2}dx \ = \ about \ 185 \ cu. \ in.$