MHB Q2:2 Where E Is Bounded By The Parabolic Cylinder

[karush]

$\text{Evaluate } $
\begin{align*}
I&=\iiint\limits_{E} x^2 e^y dV
\end{align*}
$\text{where E is bounded by the parabolic cylinder} $
\begin{align*} z&=1 - y^2 \end{align*}
$\text{and the planes
$z=0, x=1,$ and $x=-1$}\\$

 

[MarkFL]

Let's first look at the volume in question:

View attachment 7329

What order of integration makes sense to you?

 

[karush]

Let's first look at the volume in question:
What order of integration makes sense to you?

Sorta maybe??$\displaystyle\int_{0}^{1 - y^2

}\int_{-1}^{1}\int_{x_l}^{x_u} x^2 e^y \,dx \,dy \,dz$

 

[MarkFL]

Sorta maybe??$\displaystyle\int_{0}^{1 - y^2

}\int_{-1}^{1}\int_{x_l}^{x_u} x^2 e^y \,dx \,dy \,dz$

What happens when you try iterating in that order, and I assume you mean:

$$I=\int_0^{1-y^2}\int_{-1}^1\int_{-1}^1 x^2e^y\,dx\,dy\,dz$$

 

[HallsofIvy]

Sorta maybe??$\displaystyle\int_{0}^{1 - y^2

}\int_{-1}^{1}\int_{x_l}^{x_u} x^2 e^y \,dx \,dy \,dz$

Your "inside integral", with respect to z, has limits that are functions of y. So the integral will be a function of y not a number.

Also "the parabolic cylinder z= 1- y^2" does not bound a finite region. Did you mean "in the first octant"?

 

[karush]

Your "inside integral", with respect to z, has limits that are functions of y. So the integral will be a function of y not a number.

Also "the parabolic cylinder z= 1- y^2" does not bound a finite region. Did you mean "in the first octant"?

how can the "inside integral" be in respect to z?

 

[HallsofIvy]

how can the "inside integral" be in respect to z?

Yes, that should have been "outside". Thanks for pointing that out.

 

[MarkFL]

What happens when you try iterating in that order, and I assume you mean:

$$I=\int_0^{1-y^2}\int_{-1}^1\int_{-1}^1 x^2e^y\,dx\,dy\,dz$$

I was hoping you would discover on your own that with the integral written this way you would obtain a function of $y$, rather than a numeric result. Anyway, I would recommend:

$$I=\int_{-1}^1\int_{-1}^1\int_{0}^{1-y^2}x^2e^y\,dz\,dx\,dy$$

It really makes little difference as long as $z$ is not the outermost variable. Can you see any symmetries you can use to simplify the integral? This question is directed to the OP only...;)

 

[MarkFL]

This is a follow-up question related to this problem...

If done correctly, you are going to find a formula for the following useful:

$$I_n(1)-I_n(-1)=\int_{-1}^{1} x^ne^x\,dx$$ where $n\in\mathbb{N_0}$

To begin, see if you can verify:

$$I_n(x)=\int x^ne^x\,dx=n!e^x\sum_{k=0}^n\left(\frac{(-1)^k}{(n-k)!}x^{n-k}\right)$$

Hence:

$$I_n(1)-I_n(-1)=\int_{-1}^{1} x^ne^x\,dx=\frac{n!}{e}\sum_{k=0}^n\left(\frac{(-1)^ke^2+(-1)^{n+1}}{(n-k)!}\right)$$

 

[karush]

Let's first look at the volume in question:
What order of integration makes sense to you?

What graphing program did you use?

 

[MarkFL]

What graphing program did you use?

I used this site:

3D Function Grapher