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\centerline
{
\bf Dr. Z's Math251 Handout \#15.1 (2nd ed.) [Integration in Several Variables]
}

By Doron Zeilberger

{\bf Problem Type 15.1a}: Calculate the iterated integral
$$
\int_{a}^{b} \int_{c}^{d} f(x,y) \, dx \, dy \quad .
$$

{\bf Example Problem 15.1a}: Calculate the iterated integral
$$
\int_{1}^{4} \int_{0}^{2} (x+ \sqrt{y})\, dx \, dy
\quad .
$$
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} First evaluate the {\bf inner integral}.
$$
\int_{c}^{d} f(x,y) \, dx \quad .
$$
Since it is an $x$-integral, integrate with respect
to $x$, and treat $y$ as a constant. 
The answer to this step should be an expression in $y$.
Of course, once in awhile it could be just a number.
&
{\bf 1.} The inner integral is
$$
\int_{0}^{2} (x+ \sqrt{y})\, dx
$$
The anti-derivative is $x^2/2+x\sqrt{y}$, so
$$
\int_{0}^{2} (x+ \sqrt{y})\, dx= 
{{x^2} \over {2}} +x\sqrt{y} \quad \Bigl \vert_{0}^{2}
$$
$$
={{2^2-0^2} \over {2}} +(2-0)\sqrt{y}=2+2y^{1/2} \quad.
$$

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{\bf 2.} Do the {\bf outer integral} by integrating
w.r.t. $y$ the expression that you got in step 1.

&
{\bf 2.} 
$$
\int_{1}^{4} \int_{0}^{2} (x+ \sqrt{y})\, dx \, dy
$$
$$
=\int_{1}^{4} \left [ \int_{0}^{2} (x+ \sqrt{y})\, dx \right ]\, dy
$$
$$
=\int_{1}^{4}  [2+2y^{1/2}] \, dy=
2y+2{{y^{3/2}} \over {3/2}} \Bigl \vert_{1}^{4}
$$
$$
=2y+{{4} \over {3}} (\sqrt{y})^3 \Bigl \vert_{1}^{4}=
2 \cdot (4-1)+  {{4} \over {3}} [(\sqrt{4})^3 -(\sqrt{1})^3]
$$
$$
=6+  {{4} \over {3}} \cdot 7 ={{46} \over {3}}  \quad .
$$
{\bf Ans.:} $46/3$.
\cr}


{\bf Problem Type 15.1b}: Calculate the double integral
$$
\int\int_{R} f(x,y) \, dA \quad, \quad
$$
$$
R=\{(x,y) \, \vert \, a \leq x \leq b \, , \, c \leq y \leq d \, \}
\quad .
$$

{\bf Example Problem 15.1b}: 
Calculate the double integral
$$
\int\int_{R} {{xy^2} \over {x^2+1}} \, dA \quad, \quad
$$
$$
R=\{(x,y)\, \vert \, 
0 \leq x \leq 1 \, , \, -3 \leq y \leq 3 \, \} \quad.
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Convert it to an iterated integral:
$$
\int_{a}^{b}\int_{c}^{d} f(x,y) \, dy \, dx \quad, 
$$
or, if you wish
$$
\int_{c}^{d}\int_{a}^{b} f(x,y) \, dx \, dy \quad.
$$
Both are correct. 
&
{\bf 1.} 
$$
\int_{0}^{1}\int_{-3}^{3} 
{{xy^2} \over {x^2+1}}  \, dy \, dx \quad .
$$

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{\bf 2.} Evaluate the iterated intergal like in 15.1a .
&
{\bf 2.} 
$$
\int_{0}^{1}\int_{-3}^{3} 
{{xy^2} \over {x^2+1}}  \, dy \, dx \quad
$$
$$
=\int_{0}^{1} \left 
[\int_{-3}^{3} {{xy^2} \over {x^2+1}}  \, dy  \right ]
\, dx \quad .
$$
The inner integral is:
$$
\int_{-3}^{3} {{xy^2} \over {x^2+1}}   \, dy 
= {{x} \over {x^2+1}}  \int_{-3}^{3} y^2 \, dy
=\left ({{x} \over {x^2+1}} \right )
{{y^3} \over {3}} \Bigl \vert_{-3}^{3}
$$
$$
=\left ({{x} \over {x^2+1}} \right )18=
{{18x} \over {x^2+1}} \quad .
$$
Hence the iterated integral is:
$$
\int_{0}^{1} {{18x} \over {(x^2+1)}} \, dx
=9\ln(x^2+1) \Bigl \vert_{0}^{1}
$$
$$
=9[\ln(1^2+1)-\ln(0^2+1)]=9\ln 2 \quad .
$$
{\bf Ans.:} $9 \ln 2$.
\cr}

{\bf Problem Type 15.1c}: Find the volume of the
solid that lies under the plane $ax+by+cz=d$ and
above the rectangle
$$
R=\{ \, (x,y) \, \vert A_1 \leq x \leq A_2 \, , \, B_1 \leq y \leq B_2 \}
\quad .
$$
{\bf Example Problem 15.1c}: Find the volume of the
solid that lies under the plane $2x+3y+z=10$ and
above the rectangle
$$
R=\{ \, (x,y) \, \vert 0 \leq x \leq 2 \, , \, 0 \leq y \leq 1 \}
\quad .
$$

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Solve for $z$ putting it in the form
$z=f(x,y)$. Technically, you would have to check
that $f(x,y)$ is always positive above the given
region $R$ but you can trust the problem.
Set up the integral
$$
\int_R f(x,y)\, dA \quad.
$$

&
{\bf 1.} Solving for $z$ in $2x+3y+z=10$ gives
$z=10-2x-3y$, so $f(x,y)=10-2x-3y$ and the desired volume is
$$
\int_R (10-2x-3y) \, dA \quad.
$$

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{\bf 2.} Convert the area-integral to an iterated integral
either way.

&
{\bf 2.} 
$$
\int_R (10-2x-3y) \, dA \, =
\int_{0}^1 \int_{0}^{2} (10-2x-3y) \, dx \, dy
\quad .
$$
Or, if you wish
$$
\int_R (10-2x-3y) \, dA \, =
\int_{0}^2 \int_{0}^{1} (10-2x-3y) \, dy \, dx
\quad .
$$



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{\bf 3.} Compute (one of those) iterated integral(s),
like we did in 15.1a.


&
{\bf 3.} 
$$
\int_{0}^1 \int_{0}^{2} (10-2x-3y) \, dx \, dy=
\int_{0}^1 \left [ \int_{0}^{2} (10-2x-3y) \, dx \right ]\, dy
$$
$$
=\int_{0}^1 \left [ 10x-x^2-3xy \Bigl \vert_{0}^{2} \right ]\, dy=
\int_{0}^1 \left [ 10 \cdot 2-2^2-3 \cdot 2 y-0 \right ]\, dy
$$
$$
=\int_{0}^1 \left [ 16-6y \right ] \, dy
$$
$$
=16y-3y^2 \Bigl \vert_{0}^{1}=16-3=13 \quad .
$$
{\bf Ans.}: The volume is $13$.

\cr}

{\bf Problem Type 15.1d}: Find the volume of the solid
in the first octant bounded by the cylinder $z=r^2-x^2$
and the plane $y=a$.


{\bf Example Problem 15.1d}: 
Find the volume of the solid
in the first octant bounded by the cylinder $z=16-x^2$
and the plane $y=3$.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Unlike the previous problem, {\it now}
we also need to figure out the ``floor plan''
$R$. If the surface is $z=f(x,y)$ then set
$f(x,y)=0$ getting $x=\pm r$ and remember that {\it positive} octant
means $x \geq 0$, $y \geq 0$.  It follows that
$$
R=\{ (x,y) \vert 0 \leq x \leq r, 0 \leq y \leq a\} \quad .
$$
&
{\bf 1.} Setting $z=0$ gives $16 - x^2 =0$ so $x=\pm 4$ but
in the {\it positive orthant} we have $x \geq 0$ so 
the boundary of $R$ are $x=0,x=4,y=0,y=3$. So the boundary is:
$$
R=\{ (x,y) \vert 0 \leq x \leq 4, 0 \leq y \leq 3\}
\quad .
$$

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{\bf 2.} Set up the integral
$$
\int_R f(x,y)\, dA \quad ,
$$
and convert it to an iterated integral.
&
{\bf 2.} 
$$
\int_R (16-x^2) \, dA \, =
\int_{0}^{3} \int_{0}^{4} (16-x^2) \, dx \, dy \quad .
$$
\cr
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{\bf 3.} Evaluate this iterated integral.

&
{\bf 3.} 
$$
\int_{0}^{3} \int_{0}^{4} (16-x^2) \, dx \, dy=
\int_{0}^{3} \left [ \int_{0}^{4} (16-x^2) \, dx \right ]\, dy
$$
$$
= \int_{0}^{3}
\left [ 16x-{{x^3} \over {3}}  \Bigl \vert_{0}^{4} \right ]\, dy=
\int_{0}^{3}   {{128} \over {3}} \, dy
$$
$$
={{128} \over {3}} y \Bigl \vert_{0}^{3}=128 \quad .
$$
{\bf Ans.}: The volume is $128$.

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