Trying to understand how to calculate multi-dimensional integrals

Question

I've been given this assignment that includes calculating multi-dimensional (mainly in $\mathbb{R}^2$ and $\mathbb{R}^3$ ) , amusingly, with no guidance on how to.

The problem is as such:

Calculate $\int_Ee^{x^3}d\mu(x,y)$ when $E$ is a body (area/volume, not sure) in $\mathbb{R}^2$ which is blocked by the lines: $x=2 \ ,\ x=\sqrt{y}\ \ ,\ y=4 \ ,\ y=0$

I haven't tried anything except drawing the so called 'body' that is constrained within the given lines, and seeing that $E$ has no intersections with $e^{x^3}$.

I'm stuck in my thoughts, therefore any guidance/a hint towards what to do could help me progress in my understanding.

Thanks in advance!

Answer 1

The number $x$ can take any value from $0$ up to $2$. For each such $x$, the number $y$ can take any value from $0$ up to $x^2$. So, your integral is equal to $$ \int_0^2\int_0^{x^2}e^{x^3}\,\mathrm dy\,\mathrm dx=\int_0^2x^2e^{x^3}\,\mathrm dx=\left[\frac{e^{x^3}}3\right]_{x=0}^{x=2}=\frac{e^8-1}3. $$

I don't understand the reference to the line $y=2$. As far as I am concerned, that restriction is redundant.

Answer 2

After bashing my head for a healthy 60 minutes, I understood (atleast I think I did) the process:

I'm required to show that $E$ is a "Normal Reimann area", which is done by re-organizing $E$ as such: $E:=\{(x,y)\in\mathbb{R}^2:x\in[0,2],0\leq y \leq x^2 \}$

Because $[0,2]$ is a closed line in $\mathbb{R}$ therefore it's a Normal Reimann area.

And because the functions $\phi,\psi:[0,2]\rightarrow\mathbb{R}$ which are defined by $\phi(x)=0\ ,\ \psi(x)=x^2$ are continuous and $\phi\leq\psi$ for all $x\in[0,2]$ then E is normal in relation to $x$ and therefore a "Normal Reimann area" and a closed Reimann area in particular, which allows us to write $E$ as:

$E=\{(x,y)\in\mathbb{R}^2:x\in[0,2] \ , \ \phi(x)\leq y \leq \psi(x)\}$ is a Reimann area that fits $\phi,\psi$.

Since $f(x,y)=e^{x^3}$ is continuous in $\mathbb{R}^2$ and in particular in $E$, it is integrable in $E$.

Therefore, because $E$ is a Normal Reimann area and $[0,2]$ is a closed reimann area, we can use Fubini's theorem for Riemann integrals to write: $$ \int_Ee^{x^3}d\mu(x,y)=\int_D(\int_{\phi(x)}^{\psi(x)}e^{x^3}d\mu(y))d\mu(x)=\int_0^2(\int_{0}^{x^2}e^{x^3}d\mu(y))d\mu(x)=\int_0^2x^2e^{x^3}d\mu(x)=\dots=\dfrac{e^8-1}{3} $$

I hope I got the proof right, if there is a mistake or something I missed, I'm more than eager to deepen my understand in this.

Thanks in advance.