Volume of Cone in Pyramid by Slicing/Integrating This is my first time making a post on here but I am currently struggling with a problem. 
I am unsure of how to proceed with this problem as I have only really worked with volumes of revolutions rotated about an axis. 
However for this particular problem, I am asked to integrate areas by slicing a pyramid with it's top cut off with a cone bore through the middle. 
I am thinking of integrating with respect to $y$, but I don't know how to begin. Any advice helps, thank you! 
This is the specific shape of the object with dimensions in cm

 A: HINT
We have that


*

*the area of a slice for the pyramid with it's top cut off is


$$P(y)=\left(20-\frac{20+10}{20}y\right)^2$$


*

*the area of a slice for the cone is


$$C(z)=\pi \left(\frac{y}4\right)^2$$
then evaluate
$$V=\int_0^{20}(P(y)-C(y)) dy$$
A: Let $y$ denote the height above the base of a horizontal cross-section of the solid. Then we know that the area $A(y)$ of the cross-section at $y$ will have the form
\begin{equation}
A(y)=S^2(y)-\pi r^2(y)\tag{1}
\end{equation}
where $S(y)$ is the length of the side of the square at height $y$ and $r(y)$ the radius of the circle.
We see that both the length of the side of the square and the radius of the circle change linearly with $y$.
As $y$ increases from $0$ to $20$, $S$ decreases from $20$ to $10$ and $r$ increases from $0$ to $5$. 
If you were to graph $S$ with respect to $y$ you would see that the graph of $S$ would be a straight line containing points $(0,20)$ and (20,10)$. Using the point-slope formula you learned in algebra you can see that
\begin{equation}
S(y)=20-\frac{1}{2}y
\end{equation}
Likewise, the graph of $r$ versus $y$ is a straight line containing the points $(0,0)$ and $(20,5)$ so $r$ has equation
\begin{equation}
r(y)=\frac{1}{4}y
\end{equation}
Using the fact that
$$ V=\int A(y)\,dy $$
and equation (1)
we get
$$ V=\int_0^{20}\left(20-\frac{1}{2}y\right)^2-\pi\left(\frac{1}{4}y\right)^2dy $$
