Am I changing the coordinates well?

[Granger]

I have this exercise:

> $V_t=${$(x,y,z) \in \mathbb{R}^3: 1\leq x^2+y^2\leq t, 0\leq z \leq 1, y >0$}

>$F:[1,+\infty[ \rightarrow \mathbb{R}$ the function:

>$$\iiint_{V_t} \frac{e^{t(x^2+y^2)}}{x^2+y^2} \,dx\,dy\,dz$$

> Calculate $F'(4)$

Ok so the firs thing I did was to apply directly a change for cylindrical coordinates

I obtained this integral

$\int_{0}^{1} \int_{0}^{\pi} \int_{1}^{\sqrt{t}} \frac{e^{t\rho^2}}{\rho^2} \rho \,d\rho\,d\phi\,dz$

Simplifying and on this specific case ($t=4$)

$F(4)= \int_{0}^{1} \int_{0}^{\pi} \int_{1}^{2} \frac{e^{4\rho^2}}{\rho} \,d\rho\,d\phi\,dz$

For calculating $F'(4)$ I use Leibniz rule so I got the integral.

$F'(4)= \int_{0}^{1} \int_{0}^{\pi} \int_{1}^{2} e^{4\rho^2}{\rho} \,d\rho\,d\phi\,dz$

Solving the integral I got to

$F'(4) = \pi \int_{1}^{2} e^{4\rho^2}{\rho} \,d\rho = \frac{\pi}{8} (e^{16}-e^4)$

However there was some step on the integration that i did wrong because the answer should be:

$\frac{\pi}{8} (2e^{16}-e^4)$

I'm trying to figure out my mistake but I'm not getting it. Someone can please help me?

 

[I like Serena]

I have this exercise:

> $V_t=${$(x,y,z) \in \mathbb{R}^3: 1\leq x^2+y^2\leq t, 0\leq z \leq 1, y >0$}

>$F:[1,+\infty[ \rightarrow \mathbb{R}$ the function:

>$$\iiint_{V_t} \frac{e^{t(x^2+y^2)}}{x^2+y^2} \,dx\,dy\,dz$$

> Calculate $F'(4)$

Ok so the firs thing I did was to apply directly a change for cylindrical coordinates

I obtained this integral

$\int_{0}^{1} \int_{0}^{\pi} \int_{1}^{\sqrt{t}} \frac{e^{t\rho^2}}{\rho^2} \rho \,d\rho\,d\phi\,dz$

Simplifying and on this specific case ($t=4$)

$F(4)= \int_{0}^{1} \int_{0}^{\pi} \int_{1}^{2} \frac{e^{4\rho^2}}{\rho} \,d\rho\,d\phi\,dz$

For calculating $F'(4)$ I use Leibniz rule so I got the integral.

$F'(4)= \int_{0}^{1} \int_{0}^{\pi} \int_{1}^{2} e^{4\rho^2}{\rho} \,d\rho\,d\phi\,dz$

Solving the integral I got to

$F'(4) = \pi \int_{1}^{2} e^{4\rho^2}{\rho} \,d\rho = \frac{\pi}{8} (e^{16}-e^4)$

However there was some step on the integration that i did wrong because the answer should be:

$\frac{\pi}{8} (2e^{16}-e^4)$

I'm trying to figure out my mistake but I'm not getting it. Someone can please help me?

Hi GrangerObliviat! ;)

I'm afraid that Leibniz's rule has to be applied before substituting $t=4$.

And before applying it, let's simplify F(t) a bit first by changing the order of integration, so we're only left with the integral with respect to $\rho$. (Thinking)

 

[Granger]

Hi GrangerObliviat! ;)

I'm afraid that Leibniz's rule has to be applied before substituting $t=4$.

And before applying it, let's simplify F(t) a bit first by changing the order of integration, so we're only left with the integral with respect to $\rho$. (Thinking)

Hi!
I kind of did that so I ended up with $F'(t) = \frac{\pi}{2t} \int_{1}^{\sqrt{t}} 2t\rho e^{t\rho^2} \,d\rho$

And solving it and then making t=4 I get to the same thing :/

 

[I like Serena]

Hi!
I kind of did that so I ended up with $F'(t) = \frac{\pi}{2t} \int_{1}^{\sqrt{t}} 2t\rho e^{t\rho^2} \,d\rho$

And solving it and then making t=4 I get to the same thing :/

Well... there's a couple of steps missing there... (Worried)

Leibniz integral rule states that:

$$\frac{\mathrm{d}}{\mathrm{d}t} \left (\int_{a(t)}^{b(t)} f(x,t)\,\mathrm{d}x \right )= \int_{a(t)}^{b(t)}\frac{ \partial f}{ \partial t}\,\mathrm{d}x \,+\, f\big(b(t),t\big)\cdot b'(t) \,-\, f\big(a(t),t \big)\cdot a'(t)$$

For starters, what did you get for $f(\rho, t)$? (Wondering)

 

[Granger]

Well... there's a couple of steps missing there... (Worried)

Leibniz integral rule states that:

$$\frac{\mathrm{d}}{\mathrm{d}t} \left (\int_{a(t)}^{b(t)} f(x,t)\,\mathrm{d}x \right )= \int_{a(t)}^{b(t)}\frac{ \partial f}{ \partial t}\,\mathrm{d}x \,+\, f\big(b(t),t\big)\cdot b'(t) \,-\, f\big(a(t),t \big)\cdot a'(t)$$

For starters, what did you get for $f(\rho, t)$? (Wondering)

I have $f(\rho,t) = \frac {e^{t\rho^2}}{\rho^2}$

 

[I like Serena]

I have $f(\rho,t) = \frac {e^{t\rho^2}}{\rho^2}$

Let's see... we have:
$$F(t) = \int_{0}^{1} \int_{0}^{\pi} \int_{1}^{\sqrt{t}} \frac{e^{t\rho^2}}{\rho^2} \rho \,d\rho\,d\phi\,dz
= 1\cdot \pi \cdot \int_{1}^{\sqrt{t}} \frac{e^{t\rho^2}}{\rho^2} \rho\,d\rho
= \int_{1}^{\sqrt{t}} \frac{\pi e^{t\rho^2}}{\rho}\,d\rho
$$

Shouldn't that give us the following?
$$f(\rho, t) = \frac{\pi e^{t\rho^2}}{\rho}$$

I think that's what we should substitute in Leibniz integral rule.
What will that give us? (Wondering)

 

[Granger]

Yes right I misunderstood what you asked me yes that's the expression I have after all te simplification we talked about two posts ago :) But if we know apply the leibniz rule (derivating in order of t) and then integrate in order of x and then making t=4 I'm obtaining the same result :/

 

[I like Serena]

Yes right I misunderstood what you asked me yes that's the expression I have after all te simplification we talked about two posts ago :) But if we know apply the leibniz rule (derivating in order of t) and then integrate in order of x and then making t=4 I'm obtaining the same result :/

Okay... let's go through this step by step... (Thinking)

Leibniz integration rule says:
$$F'(t) = \frac{\mathrm{d}}{\mathrm{d}t} \left (\int_{a(t)}^{b(t)} f(\rho,t)\,\mathrm{d}\rho \right )= \int_{a(t)}^{b(t)}\frac{ \partial f}{ \partial t}\,\mathrm{d}\rho \,+\, f\big(b(t),t\big)\cdot b'(t) \,-\, f\big(a(t),t \big)\cdot a'(t)$$
and we have:
$$f(\rho, t) = \frac{\pi e^{t\rho^2}}{\rho},\quad a(t)=1, \quad b(t)=\sqrt t$$
So:
\begin{aligned}F'(t) &= \int_{1}^{\sqrt t}\frac{ \partial}{ \partial t}\left(\frac{\pi e^{t\rho^2}}{\rho}\right)\,\mathrm{d}\rho
\,+\,\frac{\pi e^{t(\sqrt t)^2}}{\sqrt t}\cdot \frac 1{2\sqrt t} \,-\, f\big(a(t),t \big)\cdot 0 \\
&= \int_{1}^{\sqrt t} \frac{\pi e^{t\rho^2}\cdot \rho^2}{\rho}\,\mathrm{d}\rho\,+\,
\frac{\pi e^{t^2}}{2t} \\
&= \int_{1}^{\sqrt t} \pi e^{t\rho^2} \rho\,\mathrm{d}\rho\,+\, \frac{\pi e^{t^2}}{2t} \\
&= \frac{\pi e^{t\rho^2}}{2t}\Big|_{1}^{\sqrt t}\,+\, \frac{\pi e^{t^2}}{2t} \\
&= \frac{\pi e^{t^2}}{2t} \,-\, \frac{\pi e^{t}}{2t}\,+\, \frac{\pi e^{t^2}}{2t} \\
&= \frac{\pi e^{t^2}}{t} \,-\, \frac{\pi e^{t}}{2t} \\
\end{aligned}

Did you have that as well? (Wondering)

 

[Granger]

Okay... let's go through this step by step... (Thinking)

Leibniz integration rule says:
$$F'(t) = \frac{\mathrm{d}}{\mathrm{d}t} \left (\int_{a(t)}^{b(t)} f(\rho,t)\,\mathrm{d}\rho \right )= \int_{a(t)}^{b(t)}\frac{ \partial f}{ \partial t}\,\mathrm{d}\rho \,+\, f\big(b(t),t\big)\cdot b'(t) \,-\, f\big(a(t),t \big)\cdot a'(t)$$
and we have:
$$f(\rho, t) = \frac{\pi e^{t\rho^2}}{\rho},\quad a(t)=1, \quad b(1)=\sqrt t$$
So:
\begin{aligned}F'(t) &= \int_{1}^{\sqrt t}\frac{ \partial}{ \partial t}\left(\frac{\pi e^{t\rho^2}}{\rho}\right)\,\mathrm{d}\rho
\,+\,\frac{\pi e^{t(\sqrt t)^2}}{\sqrt t}\cdot \frac 1{2\sqrt t} \,-\, f\big(a(t),t \big)\cdot 0 \\
&= \int_{1}^{\sqrt t} \frac{\pi e^{t\rho^2}\cdot \rho^2}{\rho}\,\mathrm{d}\rho\,+\,
\frac{\pi e^{t^2}}{2t} \\
&= \int_{1}^{\sqrt t} \pi e^{t\rho^2} \rho\,\mathrm{d}\rho\,+\, \frac{\pi e^{t^2}}{2t} \\
&= \frac{\pi e^{t\rho^2}}{2t}\Big|_{1}^{\sqrt t}\,+\, \frac{\pi e^{t^2}}{2t} \\
&= \frac{\pi e^{t^2}}{2t} \,-\, \frac{\pi e^{t}}{2t}\,+\, \frac{\pi e^{t^2}}{2t} \\
&= \frac{\pi e^{t^2}}{t} \,-\, \frac{\pi e^{t}}{2t} \\
\end{aligned}

Did you have that as well? (Wondering)

I realized what I was doing wrong! I did not considered that we need to the extremes of integration when they depend on t applying the Leibniz Rule..

Thank you so much!