Chapter 13, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

 a.    Assuming g is integrable and a, b, c. and d are constants,

     ${\displaystyle {\int }_c^d{\displaystyle {\int }_a^{b}g\left(x,y\right)dxdy=\left({\displaystyle {\int }_a^{b}g\left(x,y\right)dx}\right)\left({\displaystyle {\int }_c^{d}g\left(x,y\right)dy}\right).}}$

 b.     $\left\{\left(\rho ,\phi ,\theta \right):\phi =\pi /2\right\}=\left\{\left(r,\theta ,\text{}z\right):z=0\right\}=\left\{\left(x,y,z\right):z=0\right\}$

 c.    Changing the order of integration in ${\displaystyle \underset{D}{\iiint }f\left(x,y,z\right)dxdydz}$ from dx dy dz to dy dz dx also requires changing the integrand from f(x, y, z) to f(y, z, x).

d.    The transformation T: x = v, y –u maps a square in the uv-plane into a triangle in the xy-plane.

To determine

Whether the statement “ ${\displaystyle \underset{c}{\overset{d}{\int }}{\displaystyle \underset{a}{\overset{b}{\int }}g\left(x,y\right)}dxdy}=\left({\displaystyle \underset{a}{\overset{b}{\int }}g\left(x,y\right)}dx\right)\left({\displaystyle \underset{c}{\overset{d}{\int }}g\left(x,y\right)}dy\right)$, where g is integrable and a,b,c,d are constants” is true or not.

Answer to Problem 1RE

The statement is not true.

Explanation of Solution

Given:

The integrable function g and the constants a,b,c,d.

Theorem used:

Fubini’s Theorem: Let f be continuous on the rectangular region

$R=\left\{\left(x,y\right):a\le x\le b,c\le y\le d\right\}$.

The double integral of f over R may be evaluated by either of two iterated integrals:

${\displaystyle \underset{R}{\iint }f\left(x,y\right)}dA={\displaystyle \underset{c}{\overset{d}{\int }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x,y\right)}dxdy=}{\displaystyle \underset{a}{\overset{b}{\int }}{\displaystyle \underset{c}{\overset{d}{\int }}f\left(x,y\right)}dy}dx$

Reason:

Use the above mentioned Fubini’s theorem to prove this statement.

When we write ${\displaystyle \underset{c}{\overset{d}{\int }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x,y\right)}dxdy}$, it means ${\displaystyle \underset{c}{\overset{d}{\int }}\left({\displaystyle \underset{a}{\overset{b}{\int }}f\left(x,y\right)}dx\right)dy}$.

Consider the example the volume of a solid bounded by the surface $g\left(x,y\right)=xy$ over the region $R=\left\{\left(x,y\right):-1\le x\le 3,0\le y\le 2\right\}$. Simplify the left hand side of the given equation,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{-1}{\overset{3}{\int }}xy}dxdy}={\displaystyle \underset{0}{\overset{2}{\int }}\left({\displaystyle \underset{-1}{\overset{3}{\int }}xy}dx\right)dy}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}{\left(\frac{x^2}{2}y\right)}_{-1}^{3}dy}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}\left(\frac{9y}{2}-\frac{y}{2}\right)dy}\\ ={\displaystyle \underset{0}{\overset{2}{\int }}4ydy}\end{array}$

On further simplification,

$\begin{array}{c}{\displaystyle \underset{0}{\overset{2}{\int }}4ydy}={\left(\frac{4y^2}{2}\right)}_0^2\\ =\frac{16}{2}-0\\ =8\end{array}$

${\displaystyle \underset{0}{\overset{2}{\int }}4ydy}=$8 (1)

Implify the Right hand side of the give equation,

$\begin{array}{c}\left({\displaystyle \underset{a}{\overset{b}{\int }}g\left(x,y\right)}dx\right)\left({\displaystyle \underset{c}{\overset{d}{\int }}g\left(x,y\right)}dy\right)=\left({\displaystyle \underset{-1}{\overset{3}{\int }}xy}dx\right)\left({\displaystyle \underset{0}{\overset{2}{\int }}xy}dy\right)\\ ={\left(\frac{x^{2}y}{2}\right)}_{-1}^3{\left(\frac{x{y}^2}{2}\right)}_0^2\\ =\left(\frac{9y}{2}-\frac{y}{2}\right)\left(\frac{4x}{2}-0\right)\\ =\text{}\left(\frac{8y}{2}\right)\left(2x\right)\end{array}$

On further simplification $\text{}\left(\frac{8y}{2}\right)\left(2x\right)=8xy$ (2)

From the equations (1) and (2), we can find that the evaluated values are not the same.

Hence, the statement is not true.

To determine

Whether the statement

$\left\{\left(\rho ,\phi ,\theta \right):\phi =\frac{\pi }{2}\right\}=\left\{\left(r,\theta ,z\right):z=0\right\}=\left\{\left(x,y,z\right):z=0\right\}$ is true or not.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Given:

$\left\{\left(\rho ,\phi ,\theta \right):\phi =\frac{\pi }{2}\right\},\left\{\left(r,\theta ,z\right):z=0\right\},\left\{\left(x,y,z\right):z=0\right\}$

Reason:

The set $\left\{\left(\rho ,\phi ,\theta \right):\phi =\frac{\pi }{2}\right\}$ has their $\phi$ coordinate with the value $\frac{\pi }{2}$. Here $\phi$ is the angle the line to the point makes with the z-axis.

Thereby the set is same as the remaining two sets $\left\{\left(r,\theta ,z\right):z=0\right\},\left\{\left(x,y,z\right):z=0\right\}$, which are also the set of points in the xy-plane.

Hence, the statement is true.

To determine

To find: Whether the change of order of integration in ${\displaystyle \underset{D}{\iiint }f\left(x,y,z\right)}dxdydz$ from $dxdydz$ to $dydzdx$ also requires the changing the integrand from $f\left(x,y,z\right)\text{to}f\left(y,z,x\right)$ is true or not.

Answer to Problem 1RE

The statement is not true.

Explanation of Solution

Given:

The integral ${\displaystyle \underset{D}{\iiint }f\left(x,y,z\right)}dxdydz$.

Theorem used:

Let f be continuous over the region

$D=\left\{\left(x,y,z\right):a\le x\le b,g\left(x\right)\le y\le h\left(x\right),G\left(x,y\right)\le z\le H\left(x,y\right)\right\}$,

where g,h,G and H are continuous functions. Then f is integrable over D and the triple integral is evaluated as the iterated integral

$D=\left\{\left(x,y,z\right):a\le x\le b,g\left(x\right)\le y\le h\left(x\right),G\left(x,y\right)\le z\le H\left(x,y\right)\right\}$

Reason:

Consider the example,

$\begin{array}{l}{\displaystyle \underset{D}{\iiint }(2-z)dV={\displaystyle \underset{0}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}\left(2-z\right)dzdydx}}}}\\ ={\displaystyle \underset{0}{\overset{3}{\int }}\left({\displaystyle \underset{0}{\overset{2}{\int }}\left({\displaystyle \underset{0}{\overset{1}{\int }}\left(2-z\right)dz}\right)}dy\right)}dx\end{array}$

Use the above theorem to change the order of integration in the above example,

$\begin{array}{l}{\displaystyle \underset{D}{\iiint }(2-z)dV={\displaystyle \underset{0}{\overset{3}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{2}{\int }}\left(2-z\right)dydzdx}}}}\\ ={\displaystyle \underset{0}{\overset{3}{\int }}\left({\displaystyle \underset{0}{\overset{2}{\int }}\left({\displaystyle \underset{0}{\overset{1}{\int }}\left(2-z\right)dy}\right)}dz\right)}dx\end{array}$

It is observed that the change in order of integration does not alter the integrand.

Hence, the statement is not true.

To determine

Whether the transformation $T:x=v,y=-u$ maps a square in the uv-plane into a triangle in the xy-plane is true or not.

Answer to Problem 1RE

The statement is not true.

Explanation of Solution

Given:

The transformations are $T:x=v,y=-u$.

Reason:

Take the image of S in the  uv- plane is $S=\left\{\left(u,v\right):0\le u\le 1,0\le v\le 1\right\}$ be a unit square.

The uv- plane bounded by the vertices $\left(0,0\right)$, $\left(1,0\right)$, $\left(1,1\right)$ and $\left(0,1\right)$.

From $\left(0,0\right)$ to $\left(1,0\right)$, here v is fixed that is $v=0$ and the range of u varies from 0 to 1. That is, $0\le u\le 1$.

Substitute $v=0$ is in the given transformations to get $x=0$ and $y=-u$.

Therefore, xy- plane traces out the segment from $\left(0,0\right)$ to $\left(0,-1\right)$.

From $\left(0,1\right)$ to $\left(0,0\right)$, here u is fixed that is $u=0$ and the range of v varies from 0 to 1. That is, $0\le v\le 1$.

Substitute $u=0$ is in the given transformations to get $x=v\text{and}y=0$

Therefore, xy- plane traces out the segment from $\left(1,0\right)$ to $\left(0,0\right)$.

From $\left(1,0\right)$ to $\left(1,1\right)$, u is fixed that is $u=1$ and the range of v varies from 0 to 1. That is, $0\le v\le 1$.

Substitute $u=1$ is in the given transformations to get $x=v\text{and}y=-1$.

Therefore, xy- plane traces out the segment from $\left(0,-1\right)\text{to}\left(1,-1\right)$.

From $\left(1,1\right)$ to $\left(0,1\right)$, v is fixed that is $v=1$ and the range of u varies from 0 to 1. That is, $0\le u\le 1$.

Substitute $v=1$ is in the given transformations to get $x=1$ and $y=-u$.

Therefore, xy- plane traces out the segment from $\left(1,0\right)\text{to}\left(1,-1\right)$.

Thus, the image of region in xy- plane is a square with vertices $\left(0,0\right),\left(0,-1\right),\left(1,-1\right),\left(1,0\right)$

Hence, it does not maps into a triangle and thereby the statement is not true.