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\centerline
{
\bf Dr. Z's Math251 Handout \#15.3 (2nd ed.) [Triple Integrals]
}

By Doron Zeilberger

{\bf Problem Type 15.3a}: Evaluate the iterated integral
$$
\int_{a}^{b} \int_{f_1(x)}^{f_2(x)} 
\int_{g_1(x,y)}^{g_2(x,y)} F(x,y,z) \,dz \, dy \, dx \quad.
$$

{\bf Example Problem 15.3a}: 
Evaluate the iterated integral
$$
\int_{0}^{1} \int_{x}^{2x} \int_{0}^{y} 2xyz \,dz \, dy \, dx \quad.
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} You go from {\bf inside to outside}.
First isolate and compute the {\bf inner} integral
$$
\int_{g_1(x,y)}^{g_2(x,y)} F(x,y,z) \, dz  \quad,
$$
which in this problem is a $z$-integral, since
in $dz \, dy \, dx$, $dz$ comes {\bf first}.
The answer should not have $z$ in it, but in general
depends on $x$ and $y$.

&
{\bf 1.} 
$$
\int_{0}^{y} 2xyz \,dz= xyz^2 \Bigl \vert_{0}^{y}=
xy\cdot y^2-0=xy^3 \quad.
$$

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{\bf 2.} Compute the {\bf middle integral}
$$
\int_{f_1(x)}^{f_2(x)} \int_{g_1(x,y)}^{g_2(x,y)} F(x,y,z) \,dz \, dy 
\quad,
$$
using what you already know from Step 1. This should be
an expression in $x$.

&
{\bf 2.} 
$$
 \int_{x}^{2x} \int_{0}^{y} 2xyz \,dz \, dy
$$
$$
= \int_{x}^{2x} \left [ \int_{0}^{y} 2xyz \,dz \right ]\, dy
$$
from step 1, this equals:
$$
= \int_{x}^{2x} xy^3 \, dy \quad .
$$
This integral equals:
$$
x{{y^4} \over {4}} \Bigl \vert_{x}^{2x}=
x{{(2x)^4} \over {4}}-x{{(x)^4} \over {4}}=
{{15} \over {4}}x^5 \quad.
$$



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{\bf 3.} Do the outside integral, by viewing it as
$$
\int_{a}^{b} \left [ \int_{f_1(x)}^{f_2(x)} 
\int_{g_1(x,y)}^{g_2(x,y)} F(x,y,z) \,dz \, dy \, \right ]
dx \quad.
$$
and using the result from step 2.
The final answer should not depend on anything involving $x,y,z$
(i.e. is a pure number or a constant).
&
{\bf 3.} 
$$
\int_{0}^{1} \int_{x}^{2x} \int_{0}^{y} 2xyz \,dz \, dy \, dx 
$$
$$
=\int_{0}^{1} 
\left [ \int_{x}^{2x} \int_{0}^{y} 2xyz \,dz \, dy \, \right ]
dx \quad.
$$
By step 2, this is
$$
=\int_{0}^{1} {{15} \over {4}}x^5 \, dx \quad.
$$
This is a Calc I integral
$$
={{15} \over {4}} \int_{0}^{1} x^5 \, dx \quad
={{15} \over {4}} {{x^6} \over {6}}  \Bigl \vert_{0}^{1}=
{{5} \over {8}}  \quad.
$$
{\bf Ans.}: ${{5} \over {8}}$.

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{\bf Problem Type 15.3b}: 
Evaluate the triple integral
$$
\int\int\int_{E} F(x,y,z) \, dV \quad ,
$$
where
$$
E=\{ (x,y,z) \, \vert \, \, a \leq x \leq b
\, , \, f_1(x) \leq y \leq f_2(x) \, ,  \,
g_1(x,y) \leq z \leq g_2(x,y) \, \} \quad .
$$

{\bf Example Problem 15.3b}: Evaluate the triple integral
$$
\int\int\int_{E} yz \cos (x^5) \, dV \quad ,
$$
where
$$
E=\{ (x,y,z) \, \vert \, 0 \leq x \leq 1 \, , \, 0 \leq y \leq x
\, ,  \, x \leq z \leq 2x \, \} \quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Convert the volume integral into a triple iterated
interal by transcribing the limits from left to right.
$$
\int\int\int_{E} F(x,y,z) dV \quad
$$
$$
\int_{a}^{b}\int_{f_1(x)}^{f_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}
 F(x,y,z)  \, dz \, dy \, dx \quad .
$$
&
{\bf 1.} Our volume integral equals the iterated integral
$$
\int_{0}^{1}\int_{0}^{x}\int_{x}^{2x} yz\cos(x^5)  \, dz \, dy \, dx \quad .
$$

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{\bf 2.} Compute the iterated integral like we did in 15.3a.
First the inner intgeral, then the middle, then the outer.
&
{\bf 2.} The inner integral is:
$$
\int_{x}^{2x} yz \cos(x^5)  \, dz =
y \cos(x^5)  \int_{x}^{2x} z  \, dz 
$$
$$
=y \cos(x^5)  {{z^2} \over {2}} \Bigl \vert_{x}^{2x}
=y \cos(x^5)  {{(2x)^2-x^2} \over {2}} 
$$
$$
={{3} \over {2}} y x^2 \cos(x^5)   \quad .
$$
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{\bf 3.} Compute the middle integral
$$
\int_{f_1(x)}^{f_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}
 F(x,y,z)  \, dz \, dy \quad ,
$$
by writing it as
$$
\int_{f_1(x)}^{f_2(x)}
\left [ \int_{g_1(x,y)}^{g_2(x,y)}  F(x,y,z)  \, dz \right ] 
\, dy \quad ,
$$
and using the result from the previous step.

&
{\bf 3.} 
$$
\int_{0}^{x}\int_{x}^{2x} yz\cos(x^5)  \, dz \, dy \,=
\int_{0}^{x} \left [ \int_{x}^{2x} yz\cos(x^5)  \, dz \right ]\, dy \,
$$
By step 2, this equals
$$
\int_{0}^{x} {{3} \over {2}} y x^2 \cos(x^5) \, dy 
={{3} \over {2}} x^2 \cos(x^5) \int_{0}^{x}  y  \, dy
$$
$$
={{3} \over {2}} x^2 \cos(x^5) {{y^2} \over {2}} \Bigl \vert_{0}^{x}  
$$
$$
={{3} \over {2}} x^2 \cos(x^5) {{x^2-0^2} \over {2}} 
={{3} \over {4}} x^4 \cos(x^5) \quad .
$$
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{\bf 4.} Compute the outside integral
$$
\int_{a}^{b} \int_{f_1(x)}^{f_2(x)}\int_{g_1(x,y)}^{g_2(x,y)}
 F(x,y,z)  \, dz \, dy \, dx \quad ,
$$
by writing it as
$$
\int_{a}^{b} \left [\int_{f_1(x)}^{f_2(x)}
 \int_{g_1(x,y)}^{g_2(x,y)}  F(x,y,z)  \, dz
\, dy \right ] \, dx \quad ,
$$
and using the result from the previous step.
&
{\bf 4.} 
$$
\int_{0}^{1}\int_{0}^{x}\int_{x}^{2x} yz\cos(x^5)  \, dz \, dy \, dx 
$$
$$
=\int_{0}^{1}
\left [ \int_{0}^{x}\int_{x}^{2x} yz\cos(x^5)  \, dz \, dy \, \right ]
dx \quad .
$$
By step 3, this equals
$$
=\int_{0}^{1}{{3} \over {4}} x^4 \cos(x^5)\, dx 
={{3} \over {4}} \int_{0}^{1} x^4 \cos(x^5)\, dx \quad .
$$
Making the $u$-substitution $u=x^5$, gives that this equals
$$
={{3} \over {4}} \cdot {{1} \over {5}} \sin(x^5) \Bigl \vert_{0}^{1}
={{3} \over {20}}   (\sin(1) -\sin(0))=
{{3 \sin(1)} \over {20}}  \quad .
$$
{\bf Ans.}: ${{3 \sin(1)} \over {20}}$.
\cr}

{\bf Problem Type 15.3c}: 
Evaluate the triple integral
$$
\int\int\int_{E} F(x,y,z) \, dV \quad,
$$
where $E$ is bounded by the surfaces $z=g_1(x,y)$
$z=g_2(x,y)$ and possibly some planes of the form
$x=a$.

{\bf Example Problem 15.3c}:  
Evaluate the triple integral
$$
\int\int\int_{E} x^2 y^2 \, dV \quad,
$$
where $E$ is bounded by the parabolic cylinder $z=1-y^2$
and the planes $z=0$, $x=1$ and $x=-1$.


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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Express the solid region $E$ in the form
$$
E=\{(x,y,z) \vert (x,y) \in D \, , \,
$$
$$
u_1(x,y) \leq z \leq u_2(x,y) \, \} \quad ,
$$
where $D$ is the projection of $E$ on the $xy$-plane.
To find $D$, set the two surfaces equal to each other
i.e. solve $g_1(x,y)=g_2(x,y)$ and add to it the other
restictions only involving $x$ and/or $y$.
After you figure out $D$, express $E$ in the form
$$
E=\{(x,y,z) \vert a \leq x \leq b \, , g_1(x) \leq y \leq g_2(x) \, ,
$$
$$
u_1(x,y) \leq z \leq u_2(x,y) \, \} \quad .
$$

&
{\bf 1.} To get the projection $D$ solve
$1-y^2=0$, which gives $y=-1$ and $y=1$. Together
with $x=1$ and $x=-1$, the region $D$ is
$$
D=\{ \, (x,y) \, \vert -1 
\leq x \leq 1 \, , -1 \leq y \leq 1 \, \} \quad .
$$
And so
$$
E=\{(x,y,z) \vert (x,y) \in D \, , \,
0 \leq z \leq 1-y^2 \, \} \quad ,
$$
and finally (combining it with $D$), we get
$$
E=\{(x,y,z) \vert \, -1 \leq x \leq 1 \, , \, -1 \leq y \leq 1 \, , \,
0 \leq z \leq 1-y^2 \, \} \quad .
$$

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{\bf 2.}  Convert the triple integral into an iterated integral.
$$
\int_{a}^{b}\int_{g_1(x)}^{g_2(x)}\int_{u_1(x,y)}^{u_2(x,y)} 
F(x,y,z) \, dz \, dy \, dx \quad .
$$
Now evaluate that iterated integral by going from the
inside to the outside.

&
{\bf 2.} 
$$
\int_{-1}^{1}\int_{-1}^{1}\int_{0}^{1-y^2} 
x^2y^2 \, dz \, dy \, dx \quad .
$$
The inner integral is:
$$
\int_{0}^{1-y^2} x^2y^2 \, dz=
x^2y^2 \int_{0}^{1-y^2} dz=
x^2y^2(1-y^2)=x^2(y^2-y^4)
$$
The middle integral is:
$$
\int_{-1}^{1}\int_{0}^{1-y^2} 
x^2y^2 \, dz \, dy
=
\int_{-1}^{1} \left [ \int_{0}^{1-y^2} 
x^2y^2 \, dz \right ] \, dy
$$
$$
=\int_{-1}^{1} x^2 (y^2-y^4) \, dy=
x^2({{y^3} \over {3}}-{{y^5} \over {5}}) \Bigl \vert_{-1}^{1}=
{{4} \over {15}} x^2 \quad .
$$
The outer integral (i.e. the whole thing) is
$$
\int_{-1}^{1}\int_{-1}^{1}\int_{0}^{1-y^2} 
x^2y^2 \, dz \, dy \, dx \quad 
$$
$$
=\int_{-1}^{1} \left [
\int_{-1}^{1}\int_{0}^{1-y^2} 
x^2y^2 \, dz \, dy \, \right ] dx \quad . 
$$
By the previous result, this is:
$$
=\int_{-1}^{1} {{4} \over {15}}x^2 \, dx \quad 
={{4} \over {15}}{{x^3} \over {3}} \Bigl \vert_{-1}^{1}=
{{4} \over {15}} {{2} \over {3}}=
{{8} \over {45}} \quad .
$$
{\bf Ans.}:${{8} \over {45}} $.
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{\bf A Problem from a Previous Final}

Find the volume of the solid bounded by the cylinder $y=x^2$
and the planes $z=0$ and $y+z=1$. Simplify as much as you can.

{\bf Ans.} ${{8} \over {15}}$.

\end

