Triple integral: How do I include assumptions and conditions?

Question

I would like to solve the following integral: $$I=\int_{0}^{a} \int_{0}^{y} \int_{0}^{z} e^{(a-x)^{3}} d x d z d y$$ with the condition that: $$a>0$$ So I tried to input the equation like below into Mathematica, but I get error messages saying Assumptions could help.

Then I tried to do it like this but no to avail:

What am I doing wrong? Are the inputs I have mathematically incomplete, or am I doing a mistake in formatting them in Mathematica, or both??

Edit: I tried to copy the code, I hope I could do it right:

\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]\(\((
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(y\)]\((
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(z\)]
\*SuperscriptBox[\(E\), 
SuperscriptBox[\((a - 
            x)\), \(3\)]] \[DifferentialD]x)\) \[DifferentialD]z)\) \
\[DifferentialD]y\)\)

The second one:

Integrate[(\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(y\)]\(\((
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(z\)]
\*SuperscriptBox[\(E\), 
SuperscriptBox[\((a - 
           x)\), \(3\)]] \[DifferentialD]x)\) \[DifferentialD]z\)\)), \
{y, 0, a}, Assumptions -> {a > 0 && y \[Element] Reals}]

In input-converted form:

Integrate[
 Integrate[Integrate[E^(a - x)^3, {x, 0, z}], {z, 0, y}], {y, 0, a}]

The second equation:

Integrate[
 Integrate[Integrate[E^(a - x)^3, {x, 0, z}], {z, 0, y}], {y, 0, a}, 
 Assumptions -> {a > 0 && Element[y, Reals]}]

Answer 1

You should be able to use a single call to Integrate:

Integrate[Exp[(a - x)^3], {y, 0, a}, {z, 0, y}, {x, 0, z}]

Integrate[-(a*ExpIntegralE[2/3, -a^3]) + (a - z)*ExpIntegralE[2/3, -(a - z)^3], {y, 0, a}, {z, 0, y}]/3

However, it seems that Mathematica can only partially do this integral as written (note that this was evaluated in Mathematica 12.3, in earlier versions an incorrect result was returned). An alternative is to make use of Region functionality. The region of integration is:

reg = ImplicitRegion[0 < y < a && 0 < x < z && 0 < z < y, {x, y, z}];

Using this region in Integrate:

sol = Integrate[Exp[(a-x)^3], {x, y, z} ∈ reg, Assumptions -> a > 0]

1/6 (-1 + E^a^3)

Let's use NIntegrate to check for a numerical value of a:

NIntegrate[Exp[(3 - x)^3], {y, 0, 3}, {z, 0, y}, {x, 0, z}]
N[sol /. a->3]

8.86747*10^10

8.86747*10^10

Answer 2

Your error is in using the same variable name (e.g., $y$) for the variable of integration AND the integration limit.

A no-no.

This works:

    Assuming[a > 0 && {a, y, z} \[Element] Reals,
 Integrate[E^(a - x)^3, {x, 0, z}, {z, 0, y}, {y, 0, a}]]

Incidentally, you might just perform the trivial $y$ and $z$ integrals separately.