Question

Question: Evaluate the triple integral:

Short answer

The value of the given iterated integral is \(\frac{e}{2} - \frac{7}{6}\).

Step-by-step solution

The given function and Concept of Fubini’s theorem for triple integrals

The function is \(f(x,y,z) = {e^{\frac{z}{y}}}\).

The region is \(E = \{ (x,y,z)\mid y \le x \le 1,0 \le y \le 1,0 \le z \le xy\} \).

Fubini’s theorem for triple integrals:

If\(f\)is continuous on the rectangular box\(B = (a,b) \times (c,d) \times (r,s)\), then

Integrate the given integral with respect to \(z\) and apply the limit of it

Integrate the given integral with respect to \(x\) and apply the limit of it

On integrating the given integral with respect to \(x\) and applying the limit of it:

Integrate the given integral with respect to \(y\) and apply the limit of it

On integrating the given integral with respect to \(y\) and applying the limit of it:

Solve the integral \(\int_0^1 y {e^y}dy\) by using the technique of integration by parts.

\(\int u dv = uv - \int v du\).

Here, let \(u = y\), then \(du = 1 \cdot dy\) and

let \(dv = {e^y}dy\), then \(v = {e^y}\).

Substitute the values in actual integral by using Integration by Parts formula.

\(\begin{array}{l}\int_0^1 y {e^y}dy = \left( {y{e^y}} \right)_0^1 - \int_0^1 1 \cdot {e^y}dy\\ = \left( {1 \cdot {e^1} - 0 \cdot {e^0}} \right) - \left( {{e^y}} \right)_0^1\\ = e - \left( {{e^1} - {e^0}} \right)\\ = e - e + 1\\ = 1\end{array}\)

Substitute the value \(\int_0^1 y {e^y}dy = 1\) in the equation \((1)\).

Thus, the value of the given iterated integral is \(\frac{e}{2} - \frac{7}{6}\).