Calc question regarding the volume of my cheese So I've just got my cheese from the cheesemonger and he's cut it along the axis $x$,$y$, and $z$ so I can do my math homework with it.  The lengths of my cheese are:
$x$=6
$y$=4
$z$=2  
I'm asked to find the volume using calculus, because using geometry would be cheating, evidently.
The three ways I am asked to "cut up" the cheese are perpendicular to the $y$-axis (triangles), the $x$-axis (rectangles), and $z$-axis (also rectangles).
This is easy, and it turns out the volume is 24 units cubed (all three times).  E.g:
4$\int_0^6 \frac{x}{3} \mathrm{d}x = 24$
Now, I ask myself what if I want to take the other rectangle, and use that as my slice.  I know that width is 4 ($y$-axis), and the length is z$\sqrt{10}$ (pythagoras with $\sqrt{z^2 + (3z)^2}$).  But how do I find my limits of integration?
$4\sqrt{10}\int_a^b z \mathrm{d}z.$
I would need to know the distance from the $y$-axis to the top (in this photo) of the wedge.  However, I've exhausted my math knowledge here. 
I asked my calc teacher, and he told me that I find this number by "taking more calculus classes" (a good joke actually).  Since I don't have time to go bug him in his office hours, can someone explain some of the different ways to find this distance?
 A: If we were to look at the cheese with the $Y$ axis going into the board, we will be seeing a right triangle in the $X-Z$ plane right angled at the origin.
The sides of the right triangle are given by $x=0$, $z=0$ and $\frac{x}{6} + \frac{z}{2} = 1$.
A cut parallel to the "top" side of the cheese will then be the equation of the form $\frac{x}{6} + \frac{z}{2} = c$ in the $X-Z$ plane, where $c \in (0,1)$ which is parallel to the line $\frac{x}{6} + \frac{z}{2} = 1$.

Staying in the $X-Z$ plane, the perpendicular distance, say $l$, of the line $\frac{x}{6} + \frac{z}{2} = c$ from the origin can be found as follows. We want a line passing through the origin and perpendicular to $\frac{x}{6} + \frac{z}{2} = c$. So we just need to take the negative inverse of the slope of the above line and we get $z = 3x$. The intersection of these two lines is the point where the perpendicular from the origin meets the line. The point of intersection is $(\frac{3c}{5},\frac{9c}{5})$. So $l = \sqrt{(\frac{3c}{5})^2 + (\frac{9c}{5})^2} = \frac{3c \sqrt{10}}{5}$. There are also other ways to find $l$ like equating the area of the triangle calculated by two different ways. i.e. $(6c) \times (2c) = l \times (2 \sqrt{10} c) \Rightarrow l = \frac{3c \sqrt{10}}{5}$.
Now the volume element becomes $dV = 4 \times \text{Length of the line segment} \{\frac{x}{6} + \frac{z}{2} = c\} \text{bounded by the axes} \times dl$.
Note that $4$ comes from the span along the $y$ direction which is constant for all slices.
$dV = 4 \times 2 \sqrt{10} c \times \text{ }dl$ and $dl = \frac{3 \sqrt{10}}{5} dc$.
$dV = 8 \sqrt{10} \frac{3 \sqrt{10}}{5} c dc = 48 c dc$.
Hence, the volume is given by $V = \displaystyle \int_{0}^{1} 48 c dc = 24$.
