I need some help calculating the volume of a hole. The bottom of the hole is a rectangle which will have the measurements 15 by 35 meters. The depth is 8 meters. Now here is the tricky part. All four sides will slope with an angle of 27°.
It will look something like the green part of this inverted pyramid: http://thumbs.dreamstime.com/z/inverted-pyramid-segments-shows-hierarchy-24615808.jpg
The difference though is that if I extend the sides it will not be a pyramid but more lika a tent with shorter "roof" than bottom. (couldn't find a good image of exact shape)
My first thought was that I could use integration but my experience with them are limited. The problems I've solved has been quite straight forward. I'm guessing that I need to start with the bottom area and integrate it piece by piece (dx). What I think I am lacking is the ratio between each piece to make that integration. How do I find that?
The other option could be finding where the lines intersect at the top of the triangle and then subtract the top part. The problem is that it is not a pyramid so I just don't know where to start.
Anyone that could give me a little bit of help with this one?
(I have english as a second language so some words might be are wrong.)
We begin by finding the dimensions of the top face of the frustum (the geometric name of the hole you described). We see that both the top width and top breadth are $\frac{16}{\tan(27^\circ)}$ larger than the frustum's bottom width and bottom breadth. Thus, the top has dimensions of $15 + \frac{16}{\tan(27^\circ)}$ by $35 + \frac{16}{\tan(27^\circ)}.$
Now we wish to find the height of the pyramid that the frustum shares a top face with. Using the values we just found, we can see that this pyramid's height is $8 + \frac{15\tan(27^\circ)}{2}.$ We can now subtract the smaller pyramid from the larger one to find the volume of the solid. The smaller pyramid has volume $V_{s} = (\frac{1}{3})(15)(35)(\frac{15\tan(27^\circ)}{2}).$ The larger pyramid has volume $V_{l} = (\frac{1}{3})(15 + \frac{16}{\tan(27*)})(35 + \frac{16}{\tan(27^\circ)})(8 + \frac{15\tan(27^\circ)}{2}).$
The volume of the frustum (hole) is $V_{l} - V_{s},$ which is approximately $11472$ cubic meters.