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\centerline
{
\bf Solutions to the ``QUIZ'' for Lecture 14
}
\bigskip

{\bf 1.} 
Evaluate the iterated integral
$$
\int_{0}^{1} \int_{x}^{3x} \int_{0}^{y} x^2yz \,dz \, dy \, dx \quad.
$$

{\bf Sol.} We first do the {\bf inner} integral
$$
 \int_{0}^{y} x^2y z \,dz =  x^2y \int_{0}^{y}  z \,dz = x^2 y \left ( {{z^2} \over {2}} \Bigl \vert_{0}^{y} \right )
= x^2 y \left ( {{y^2} \over {2}} - {{0^2} \over {2}}  \right )
= {{x^2y^3} \over {2}} 
$$
We next do the {\bf middle} integral
$$
\int_{x}^{3x} {{x^2y^3} \over {2}} \, dy 
=
{{x^2} \over {2}} \int_{x}^{3x}  y^3 \, dy 
=
{{x^2} \over {2}} \left (  {{y^4} \over {4}} \Bigl \vert_{x}^{3x} \right )
=
{{x^2} \over {2}} \left (  {{(3x)^4-x^4} \over {4}} \right )=
$$
$$
{{x^2} \over {2}} \left (  {{81x^4-x^4} \over {4}} \right )=
{{x^2} \over {2}} \left (  {{80x^4} \over {4}} \right )
=
10x^6
$$
Finall, we do the {\bf outer} integral
$$
\int_{0}^{1} 10x^6 \, dx = 10 {{x^7} \over {7}} \Bigl \vert_{0}^{1}={{10} \over {7}} \quad.
$$

{\bf Ans.}: ${{10} \over {7}}$ (type number).

{\bf Comment}: Most people knew how to do it, but only about \%60 got the right answer.
Many people messed up the arithmetic.

{\bf 2.} 
Evaluate the triple integral
$$
\int\int\int_{E} yz \ln (x^5) \, dV \quad ,
$$
where
$$
E=\{ (x,y,z) \, \vert \, 0 \leq x \leq 1 \, , \, 0 \leq y \leq x
\, ,  \, 2x \leq z \leq 3x \, \} \quad .
$$

{\bf Sol.} We first {\bf set-up} the {\bf volume integral} as an {\bf iterated integral},
following the region $E$.
$$
\int_0^1 \int_0^x \int_{2x}^{3x} yz \ln (x^5) \, \, dz \, dy \, dx  \quad .
$$

We first do the {\bf inner integral}:
$$
 \int_{2x}^{3x} yz \ln (x^5) \, \, dz 
=
y \ln (x^5) \int_{2x}^{3x} z  \, \, dz =
y \ln (x^5) {{z^2} \over {2}} \Bigl \vert_{2x}^{3x} =
$$
$$
y \ln (x^5) {{(3x)^2-(2x)^2} \over {2}}=
y \ln (x^5) {{9x^2-4x^2} \over {2}}=
y \ln (x^5) {{5x^2} \over {2}} \quad .
$$

We next do the {\bf middle} integral:
$$
\int_0^x  y \ln (x^5) {{5x^2} \over {2}} \, dy 
=\ln (x^5) {{5x^2} \over {2}} \int_0^x y  \, dy 
=\ln (x^5) {{5x^2} \over {2}} \left (  {{y^2} \over {2}} \Bigl \vert_{0}^{x} \right )
=\ln (x^5) {{5x^2} \over {2}} \cdot  {{x^2} \over {2}}
=\ln (x^5) {{5x^4} \over {4}} \quad
$$

We finally do the {\bf outer integral}
$$
\int_{0}^{1} \ln (x^5) {{5x^4} \over {4}} \, dx
$$
Doing the {\bf substition} $z=x^5$ we get $dz=5x^4 \, dx$, and when $x=0,z=0$, and when $x=1,z=1$.
So we have
$$
\int_{0}^{1} ln (z) {{1} \over {4}} \, dz=
{{1} \over {4}} \int_{0}^{1} ln (z)  \, dz \quad .
$$
By {\bf integration by parts} 
$$
\int \ln (z)  \, dz = \int_{0}^{1} z' ln (z)  \, dz = z \ln z -  \int z (ln (z))' \, dz
= z \ln z - \int z \cdot (1/z) \, dz 
= z \ln z - \int \, dz = z \ln z -z  \quad .
$$
So we have
$$
{{1} \over {4}} \int_{0}^{1} ln (z)  \, dz = {{1} \over {4}} (z \ln z -z) \Bigl \vert_{0}^{1}
=
{{1} \over {4}} [ (1 \ln 1 -1)  - ( 0 \ln 0 -0 ) ] \quad=
{{1} \over {4}} [ (1 \cdot 0 -1)  - (0 -0 ) ]=
-{{1} \over {4}}  \quad
$$

{\bf Ans.} $-{{1} \over {4}}$ (type number).

{\bf Comments}: 1. Strictly speaking $0 \ln 0$ is nonsense, since $\ln 0$ is undefined.
What I mean by ``$0 \ln 0$'' is $\lim_{x \rightarrow 0^{+}} x \ln x$ that is well-known to
be $0$ (or use L'H\^opital in the form ${{\ln x} \over {x^{-1}}}$) .

2. Very few people finished it, since I didn't allow enough time for this hard and
long problem.

3. I often tell you to {\bf simplify before you integrate}. In this problem, an obvious simplification is
$\ln x^5= 5 \ln x$, but in this case it does not make things any simpler (in fact, it makes it slightly more complicated).



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