Chapter 16, 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, ${\displaystyle {\int }_c^d{\displaystyle {\int }_a^{b}g(x,y)}}$ dx dy = $\left({\displaystyle {\int }_a^{b}g(x,y)dx}\right)\left({\displaystyle {\int }_c^{d}g(x,y)dy}\right).$

b. The spherical equation φ = π/2, the cylindrical equation z = 0, and the rectangular equation z = 0 all describe the same set of points.

c. Changing the order of integration in ${\displaystyle \underset{D}{\iiint }f(x,y,z)}$ dx dy dz from dx dy dz to dy dz dx requires also 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 to a triangle in the xy-plane.

To determine

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

Answer to Problem 1RE

The statement is false.

Explanation of Solution

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:

${\displaystyle \underset{R}{\iint }f\left(x,y\right)}dA={\displaystyle {\int }_c^d{\displaystyle {\int }_a^{b}f\left(x,y\right)}dx}dy={\displaystyle {\int }_a^b{\displaystyle {\int }_c^{d}f\left(x,y\right)}dy}dx$

Description:

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

Use the Fubini’s theorem to prove or disprove the given statement.

The integral expression ${\displaystyle \underset{c}{\overset{d}{\int }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x,y\right)}dxdy}$ can be written as ${\displaystyle \underset{c}{\overset{d}{\int }}\left({\displaystyle \underset{a}{\overset{b}{\int }}f\left(x,y\right)}dx\right)dy}$.

Consider the example of the volume of a solid bounded by the surface $g\left(x,y\right)=xy$ over the region $R=\left\{\left(x,y\right):-1\le x\le 3,0\le y\le 2\right\}$. Thus, the integral will be ${\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{-1}{\overset{3}{\int }}xy}dxdy}=\left({\displaystyle \underset{a}{\overset{b}{\int }}g\left(x,y\right)}dx\right)\left({\displaystyle \underset{c}{\overset{d}{\int }}g\left(x,y\right)}dy\right)$.

Simplify the left hand side of the equation as follows.

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

On further simplification,

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

That is, ${\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{-1}{\overset{3}{\int }}xy}dxdy}=8$.                                                                                           (1)

Simplify the Right hand side of the equation as follows.

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

On further simplification,

$\left({\displaystyle \underset{a}{\overset{b}{\int }}g\left(x,y\right)}dx\right)\left({\displaystyle \underset{c}{\overset{d}{\int }}g\left(x,y\right)}dy\right)=8xy$ (2)

From the equations (1) and (2), the evaluated values are not the same.

Hence, the statement is false.

To determine

Whether the statement “The spherical equation $\phi =\frac{\pi }{2}$, the cylindrical equation $z=0$ and the rectangular equation $z=0$ all describe the same set of points” is true or not.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

The set of sphere for $\phi =\frac{\pi }{2}$ can be written as $\left\{\left(\rho ,\phi ,\theta \right):\phi =\frac{\pi }{2}\right\}$, where the sphere has their $\phi$ coordinate with the value $\frac{\pi }{2}$.

Here, $\phi$ is the angle made by the line at the point with the z-axis. That is, this set is the set of points in the xy-plane.

The set of cylinder for $z=0$ can be written as $\left\{\left(r,\theta ,z\right):z=0\right\},\left\{\left(x,y,z\right):z=0\right\}$.

Thereby the set $\left\{\left(r,\theta ,z\right):z=0\right\},\left\{\left(x,y,z\right):z=0\right\}$ also represent the set of points in the xy-plane.

Thus, the spherical equation $\phi =\frac{\pi }{2}$, the cylindrical equation $z=0$ and the rectangular equation $z=0$ all describe the same set of points.

Hence, the statement is true.

To determine

Whether the statement “Changing the order of integration in ${\displaystyle \underset{D}{\iiint }f\left(x,y,z\right)}dxdydz$ from $dxdydz$ to $dydzdx$ requires also 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 false.

Explanation of Solution

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:

${\displaystyle \underset{D}{\iiint }f\left(x,y,z\right)}dV={\displaystyle \underset{a}{\overset{b}{\int }}{\displaystyle \underset{g\left(x\right)}{\overset{h\left(x\right)}{\int }}{\displaystyle \underset{G\left(x,y\right)}{\overset{H\left(x,y\right)}{\int }}f\left(x,y,z\right)}}}dzdydx$

Description:

Consider the example,

$\begin{array}{c}{\displaystyle \underset{D}{\iiint }(2-z)}dV={\displaystyle {\int }_0^3{\displaystyle {\int }_0^2{\displaystyle {\int }_0^1\left(2-z\right)}dz}dy}dx\\ ={\displaystyle \underset{0}{\overset{3}{\int }}\left({\displaystyle {\int }_0^2\left({\displaystyle {\int }_0^1\left(2-z\right)}dz\right)}dy\right)}dx\end{array}$

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

$\begin{array}{c}{\displaystyle \underset{D}{\iiint }(2-z)}dV={\displaystyle {\int }_0^3{\displaystyle {\int }_0^1{\displaystyle {\int }_0^2\left(2-z\right)}dy}dz}dx\\ ={\displaystyle \underset{0}{\overset{3}{\int }}\left({\displaystyle {\int }_0^1\left({\displaystyle {\int }_0^2\left(2-z\right)}dy\right)}dz\right)}dx\end{array}$

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

Hence, the statement is false.

To determine

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

Answer to Problem 1RE

The statement is false.

Explanation of Solution

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

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

The uv-plane is 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 $\left(1,0\right)$, 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 $v=0$ in the 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 $\left(0,0\right)$, 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 $u=0$ in the 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 $u=1$ in the 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 $v=1$ in the transformations to obtain $x=1$ 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 $\left(0,0\right),\left(0,-1\right),\left(1,-1\right)\text{ and}\left(1,0\right)$.

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