Q. 48 - Calculus

Question

In Exercises 45–52, rewrite the indicated integral with the specified order of integration. 

Exercise 38 with the order dx dy dz.

Step-by-Step Solution

Answer

The given indicated integral with the specified order of integration is,

${\int }_0^4{\int }_0^2{\int }_0^{3z/4}f(x,y,z)dxdydz$

Step 1. Given Information.

We are given,

${\int }_0^2{\int }_0^3{\int }_{4x/3}^{4}f(x,y,z)dzdxdy$

Step 2. Rewriting the integral.

We have to write the given integral in order of dx dy dz. 

So, the limit of the integration will vary as follows:

The limit of x alters from $0 \text{to} 3$

The limit of y alters from $0 \text{to} 2$

The limit of z alters from $0 \text{to} \frac{4x}{3}$

When $z=\frac{4x}{3}$,

$x=\frac{3z}{4}$

Combining the above limits, x varies from $0$ to $\frac{3z}{4}$.

Step 3. Rewriting the integral.

The limits of y varies from $0$ to $2$. There will be no change.

Calculate the limits of z.

When $x=0$,

$$\begin{gathered} z=\frac{4x}{3} \\ =\frac{{\displaystyle 4\left(0\right)}}{{\displaystyle 3}} \\ =0 \end{gathered}$$

When $x=3$,

$$\begin{gathered} z=\frac{4x}{3} \\ =\frac{{\displaystyle 4\left(3\right)}}{{\displaystyle 3}} \\ =4 \end{gathered}$$

Hence the indicated integral with the specified order of integration will be,

${\int }_0^2{\int }_0^3{\int }_{4x/3}^{4}f(x,y,z)dzdxdy={\int }_0^4{\int }_0^2{\int }_0^{3z/4}f(x,y,z)dxdydz\text{. }$