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Let $$E=\{(x,y,z)\ |\ 1\leq x\leq 2 , \sqrt{x}<y<x , 0\leq z\leq xy \}$$ and $$\pi = E \cap \{(x,y,z) \ |\ x=2 \}.$$

Find the area of $\pi$ and the volume of $E$.

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No shouting, please. – lhf Nov 27 '12 at 0:06
I have typeset your question into LaTeX. Please verify that I have done so correctly. – Neal Nov 27 '12 at 0:11
It is correct. Thanks – Paul DeJong Nov 27 '12 at 0:24

This is a basic exercise in multiple integration. We have been given the boundaries we need for calculating the volume of $E$: $$ V(E) = \int_1^2\int_\sqrt{x}^x\int_0^{xy}1\,dz\,dy\,dx = \int_1^2\int_\sqrt{x}^xxy\,dy\,dx = \int_1^2\frac{1}{2}(x^3-x^2)\,dx = \frac{17}{24} $$ The basic idea of integration is that you can calculate an area by "adding together" the area of the infinitely many, infinitely small cross sectional lines, or calculate the volume of something by adding together the "volumes" of all cross sections. Let's look at it piece by piece.

The outermost layer of the integration is $$ \text{Volume of $E$} = \int_1^2 \{\text{Area of cross section of $E$ at $x$}\}\cdot dx $$ Here "Area of cross section" is just that, and while not strictly proper math, you might think of $dx$ as an infinitely small width. Thus the product of the two would be an infinitely small volume, and the integration sign is just a symbol signifying that we add them all up. (This paragraph might get some people heated up, but that doesn't change the fact that thinking this way yields the right answer, and makes intuitive sense for people as they learn the basics.)

Now, given a valid point $x_0$ on the $x$-axis, we can tell exactly what interval of the $y$-axis is valid. That is $[\sqrt{x_0}, x_0]$. Thus the cross section area, given a specific $x$ can be calculated as $$ \text{Area of cross section} = \int_\sqrt{x}^x\{\text{Length of line segment contained in $E$ given $x$ and $y$}\}\cdot dy $$ by the same reasoning. And the length of the line segment is calculated by integrating the constant function $f(x, y, z) = 1$ along the length of the segment. Given valid $x_1$ and a $y_1$, the $z$-interval for this segment in $[0, x_1y_1]$, so we integrate over that length: $$ \text{Length of line segment} = \int_0^{xy}1\,dz $$ Now, you might be scared off by the limits of the intervals so far, but don't be! The integral above is calculated just like any other. We have $\int 1 \, dz = z + C$, so we just calculate the difference $$ \int_0^{xy}1\,dz= xy - 0 = xy $$ as if $xy$ were a normal number (in fact you can think of it as though it is a normal number, only you don't know which one).

Now that we know that the length of a line segment given $x$ and $y$ is $xy$, we can calculate the cross section area given $x$: $$ \text{Area of cross section} = \int_\sqrt{x}^x xy\, dy $$ We have $\int xy\,dy = \frac{1}{2}xy^2 + C$, so the integral above becomes (once again using $\sqrt{x}$ and $x$ as plain numbers, which we can do since we're integrating with respect to $y$): $$ \int_\sqrt{x}^x xy\, dy = \frac{1}{2}x\cdot x^2 - \frac{1}{2}x\sqrt{x}^2 = \frac{1}{2}(x^3 - x^2) $$ Now the last integral is straight forward: $$ \text{Volume of $E$} = \int_1^2 \frac{1}{2}(x^3 - x^2) \, dx = \frac{1}{2}(\frac{1}{4}\cdot 2^4 - \frac{1}{3}\cdot 2^3) - \frac{1}{2}(\frac{1}{4}\cdot 1^4 - \frac{1}{3}\cdot 1^3) = \frac{17}{24} $$ and we're done.

The area of $\pi$ is just the area of the cross section at $x = 2$, so going by the reasoning above, we get $$ \int_\sqrt{2}^2\int_0^{2y}1\,dz\,dy = \int_\sqrt{2}^22y\,dy = 2 $$

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Thank you very much. That is a great explanation for whats going on. I appreciate it. – Paul DeJong Nov 27 '12 at 1:12

$E$ is a volume of $\mathbb R^3$. All points in it must satisfy all six inequalities. As an example, for $x=1.44$ (which satisfies the first two), we must have $1.2 \lt y \lt 1.44$. Choosing $y=1.3$, we must have $0 \le z \le 1.56$, so the segment from $(1.44,1.3,0)$ to $(1.44,1.3,1.56)$ is part of the volume.

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Ok thanks. How do you find the height of the solid? – Paul DeJong Nov 27 '12 at 0:21
@PaulDeJong: If by height you mean the range in $z$, I multiplied $x$ and $y$ for the maximum. – Ross Millikan Nov 27 '12 at 1:48

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