MHB 232.15.4.46 Change the Integral order then evaluate

[karush]

$\tiny{232.15.4.46}$
$\textsf{Change the order then evaluate}$
\begin{align*}\displaystyle
I&=\int_{0}^{1}\int_{0}^{2}\int_{2y}^{4}
\frac{5\cos(x^2)}{2z}
\, dx \, dy \, dz
\end{align*}

ok I presume the change that should be made is...
altho I don't know what represents x or y

\begin{align*}\displaystyle
I&=\int_{0}^{1} \int_{2y}^{4} \int_{0}^{2}
\frac{5\cos(x^2)}{2z}
\, dx \, dy \, dz
\end{align*}also I wanted to plot this on W|A but didn't know the para metric for it
\end{align*}

 

[GJA]

Hi karush,

By writing

$$I = \int_{0}^{1}\frac{5}{2z}\left[\int_{0}^{2}\int_{2y}^{4}\cos(x^{2})dx\,dy \right]dz,$$

hopefully it makes sense that the change of integration order needed pertains only to the $x$ and $y$ variables in this example (generally speaking, this need not be the case). That in mind, the term in the brackets is an integral over a region, say $A$, in $\mathbb{R}^{2}$ described by

$$A=\{(x,y): 0\leq y\leq 2~~~\&~~~ 2y\leq x\leq 4\}.$$

Can you determine what shape this region is? Once you do, can you then find a bound for $y$ in terms of $x$ on this region (as opposed to the one you currently have, which is a bound for $x$ in terms of $y$)? To help you do this, imagine drawing a vertical line anywhere that passes from the bottom of this region to the top of the region -- the new lower limit will be the equation for $y$ in terms of $x$ that corresponds to the first "curve" your vertical line passes through as it enters $A,$ and the new upper limit will be the equation for $y$ in terms of $x$ that corresponds to the second "curve" your vertical line passes through as it exits $A$.

Secondly, you need new bounds for $x$ itself in terms of fixed numbers (as opposed to the ones you currently have for $y$, which are $0$ and $2$). If you can achieve all of this you can then perform the integration in brackets as

$$\int_{?}^{?}\int_{?}^{?}\cos(x^{2})\,dy\,dx.$$

Hopefully this helps and isn't too confusing as to what is meant. Let me know if you have any questions. Good luck!

 

[karush]

Secondly, you need new bounds for $x$ itself in terms of fixed numbers (as opposed to the ones you currently have for $y$, which are $0$ and $2$). If you can achieve all of this you can then perform the integration in brackets as

$$\int_{?}^{?}\int_{?}^{?}\cos(x^{2})\,dy\,dx.$$

Hopefully this helps and isn't too confusing as to what is meant. Let me know if you have any questions. Good luck!

$$\int_{1}^{4}\int_{0}^{1}\cos(x^{2})\,dy\,dx.$$

ok I am assuming we can make adjustment with $0$ to $1$ from $dz$
but don't know where the $\frac{5}{2z}$ fits in

knowing that the dz will be same at as long as it is 1 we can place it anywhere

sorry think I'm lost on this

 

[HallsofIvy]

$\tiny{232.15.4.46}$
$\textsf{Change the order then evaluate}$
\begin{align*}\displaystyle
I&=\int_{0}^{1}\int_{0}^{2}\int_{2y}^{4}
\frac{5\cos(x^2)}{2z}
\, dx \, dy \, dz
\end{align*}

ok I presume the change that should be made is...
altho I don't know what represents x or y

\begin{align*}\displaystyle
I&=\int_{0}^{1} \int_{2y}^{4} \int_{0}^{2}
\frac{5\cos(x^2)}{2z}
\, dx \, dy \, dz
\end{align*}

This would have the second integral, with respect to y, going from 2y to 4. That's impossible! You cannot have the limits of integration, with respect to a variable, depending upon that variable.

Fortunately, the limits of the integration with respect to z do not depend upon x or y and the integrations with respect to x and y do not depend upon z so we can treat this as just a two dimensional, xy, problem. In the original integration, the integral with respect to y has limits of integration 0 and 2. On an xy graph, draw vertical lines y= 0 and y= 4. The limits of integration for the integral with respect to x has lower limit 2y and upper limit 2. Draw the vertical line x= 2 and the line x= 2y which is the same as y= x/2. That line crosses y= 0 at (0, 0) and y= 4 at (2, 4). The region of integration is the triangle with vertices (0, 0), (2, 0), and (2, 4).

To change the order of integration, so that we are integrating with respect to x first and then y: the second integral, with respect to y, must not depend upon x- the limits of integration must be number. To cover the entire region, y must go from 0 to 4. And, for each y, x must go from 0 up to the line x= y/2. The integral must be
[math]\int_0^1\int_0^4\int_0^{y/2} \frac{5 cos(x^2)}{2z}dxdydz= \frac{5}{2}\left(\int_0^1\frac{dz}{z}\right)\left(\int_0^4\int_0^{2x} cos(x^2) dy dx\right)[/math].

also I wanted to plot this on W|A but didn't know the para metric for it
\end{align*}