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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?

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    The volume is not $4 \sqrt{10} \int_{a}^{b} z dz$. since your integration now is not along z but along a line perpendicular to $\frac{x}{6} + \frac{z}{2} = 1$.2011-02-10
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    Very interesting. I'm reading your post and gears are turning now :) .2011-02-10
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    Side note: when in math will I learn how to solve this? I.e. will this be something I can do easily after I finish Calc 4? Diff eq?2011-02-10
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    I dont know. I did my high-school and undergrad in a different country and I dont know how it works here. We did this in our XI which is equivalent to junior year of high school.2011-02-10
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    @Stephano: the way it's done in Sivaram's answer really only involves early high-school-level geometry (albeit a creative application of it), assuming you're in the US. I guess you could also do it using vector calculus, but of course I don't know what number of calculus class that corresponds to at your school. (It does come before differential equations)2011-02-10
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    @David thanks David. my high school math is a wee bit fuzzy. i should really brush up on my trig and geometry.2011-02-10

1 Answers 1

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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 $\frac{x}{6} + \frac{z}{2} = 1$.

enter image description here

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 $V = \displaystyle \int_{0}^{1} 48 c dc = 24$.

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    +1 for understanding my crazy question. I'll mark as correct as soon as I understand all of this... brain... hurting... :)2011-02-10
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    wow. what program can make that awesome diagram? i used photoshop as I'm clearly not a math guy.2011-02-10
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    I have added a figure to help your understanding.2011-02-10
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    I use grapher which is plotting software on osx.2011-02-10
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    Mmm, grapher. Fun trick, thanks.2011-02-10
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    Ok, the only part I'm not 100% clear on is where your limits of integration came from. I'm guessing it is the length of the line perpendicular to $\frac{x}{6}+\frac{z}{2}$=12011-02-10
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    @Stephano: Right. These are essentially limits on $l$. $l$ goes from $0$ to the perpendicular distance to the top of the wedge i.e $l_{top} = \frac{3 \sqrt{10}}{5}$ equivalently, what this means is $c$ goes from $0$ to $1$, since $l = \frac{3 \sqrt{10}}{5} c$.2011-02-10
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    Awesome. Thank you for all your help. Well earned answer.2011-02-10