Integrations

**geton** · December 25th 2008, 08:05 PM

1. The curve C has parametric equation x = sin t, y = sin 2t, 0 ≤ t ≤ pi/2.

If the region is revolved through $2 \pi$ radians about the x-axis, find the volume of the solid formed.
-----------------------
So Volume = $\pi \int y^2 dx = \pi \int y^2 \frac {dx}{dt}dt$

$= \pi \int_{0}^{\frac{\pi}{2}} sin^2 2t \,cos\,t\, dt$

Then how could I integrate this?
-----------------------

2. Find a general solution of the following differential equation.

$e^{x+y} \frac{dy}{dx} = x(2 + e^y)$
-----------------------

So far I did,

$\int \frac{e^y}{2+e^y} dy = \int \frac{x}{e^x} dx$

$\int (1 - \frac{2}{2 + e^y})dy = \int x\, e^{-1}dx$

$y - 2 \,ln|2+e^y| = -xe^{-x} - e^{-x} + C$

But this is not right ans. The right ans is $ln|2+e^y| = -xe^{-x} - e^{-x} + C$

So where did I wrong?
-----------------------

**Danny** · December 25th 2008, 08:24 PM

Originally Posted by geton

1. The curve C has parametric equation x = sin t, y = sin 2t, 0 ≤ t ≤ pi/2.

If the region is revolved through $2 \pi$ radians about the x-axis, find the volume of the solid formed.
-----------------------
So Volume = $\pi \int y^2 dx = \pi \int y^2 \frac {dx}{dt}dt$

$= \pi \int_{0}^{\frac{\pi}{2}} sin^2 2t \,cos\,t\, dt$

Then how could I integrate this?
-----------------------

2. Find a general solution of the following differential equation.

$e^{x+y} \frac{dy}{dx} = x(2 + e^y)$
-----------------------

So far I did,

$\int \frac{e^y}{2+e^y} dy = \int \frac{x}{e^x} dx$

$\int (1 - \frac{2}{2 + e^y})dy = \int x\, e^{-1}dx$

$y - 2 \,ln|2+e^y| = -xe^{-x} - e^{-x} + C$

But this is not right ans. The right ans is $ln|2+e^y| = -xe^{-x} - e^{-x} + C$

So where did I wrong?
-----------------------

Q1 Expand

$sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t$ and use a substitution $u = cos\,t\,$

Q2

$\int \frac{e^y}{2+e^y} dy \ne y - 2 \,ln|2+e^y|$

If $u = 2+e^y$ then $du = e^y\, dy$ so $\int \frac{e^y}{2+e^y} dy = \int \frac{du}{u}$

which easily interates giving

$\int \frac{e^y}{2+e^y} dy = \,ln|2+e^y| + c$

**tsal15** · December 25th 2008, 08:45 PM

Originally Posted by danny arrigo

Q1 Expand

$sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t$ and use a substitution $u = cos\,t\,$

Q2

$\int \frac{e^y}{2+e^y} dy \ne y - 2 \,ln|2+e^y|$

If $u = 2+e^y$ then $du = e^y\, dy$ so $\int \frac{e^y}{2+e^y} dy = \int \frac{du}{u}$

which easily interates giving

$\int \frac{e^y}{2+e^y} dy = \,ln|2+e^y| + c$

Forgive my ignorance but how did you guys seperate the $e^{x+y}$ term?

Thanks

**Danny** · December 25th 2008, 08:48 PM

Originally Posted by tsal15

Forgive my ignorance but how did you guys seperate the $e^{x+y}$ term?

Thanks

One of the properties of exponents

$e^{x+y} = e^x \, e^y$

**Mathstud28** · December 25th 2008, 08:48 PM

Originally Posted by geton

2. Find a general solution of the following differential equation.

$e^{x+y} \frac{dy}{dx} = x(2 + e^y)$

As you noted $\begin{aligned}e^{x+y}\frac{dy}{dx}=x(2+e^y)&\impl ies \frac{e^y}{2+e^y}dy=\frac{x}{e^x}dx\\ &\implies\int\frac{e^y}{2+e^y}dy=\int\frac{x}{e^x} dx\\ &\implies \color{red}{\ln(2+e^y)}=-\left(\frac{x}{e^x}+\frac{1}{e^x}\right)+C\\ &\implies y=\ln\left(\exp\left(-\left\{\frac{x}{e^x}+\frac{1}{e^x}\right\}+C\right )-2\right) \end{aligned}$

The problem is that if you let $\varphi\mapsto 2+e^y$ in your integral you get $d\varphi=e^y~dy$ so you should have $\int\frac{e^y}{2+e^y}dy\stackrel{\varphi=2+e^y}{\l ongmapsto}\int\frac{d\varphi}{\varphi}$

**geton** · December 25th 2008, 09:42 PM

Originally Posted by danny arrigo

Q1 Expand

$sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t$ and use a substitution $u = cos\,t\,$

Ok,

Let $u = cos\, t$
$\frac{du}{dt} = - sin\,t$

Then?

Could you explain how to do this?

**geton** · December 25th 2008, 10:00 PM

Originally Posted by geton

Ok,

Let $u = cos\, t$
$\frac{du}{dt} = - sin\,t$

Then?

Could you explain how to do this?

So far I thought

$\int 4 \, sin^2t \,cos^3 t \,dt$
$= - 4 \int u^3 \sqrt{1-u^2}\,du$

Is it the right way? Then?

**mr fantastic** · December 25th 2008, 10:22 PM

Originally Posted by geton

So far I thought

∫4 sin^2 t cos^3 t dt
= 4∫ (1-cos^2 t)cos^3 t dt
= 4∫ (cos^3 t - cos^5 t) dt

Is it the right way? Then?

No.

First you note that $4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t$ .

Then you make the substitution ${\color{red}u = \sin t}$ (NOT $u = \cos t$ which was no doubt a typo by danny). Note that $dt = \frac{du}{\cos t}$ .

**mr fantastic** · December 25th 2008, 10:29 PM

Originally Posted by Mathstud28

As you noted $\begin{aligned}e^{x+y}\frac{dy}{dx}=x(2+e^y)&\impl ies \frac{e^y}{2+e^y}dy=\frac{x}{e^x}dx\\ &\implies\int\frac{e^y}{2+e^y}dy=\int\frac{x}{e^x} dx\\ &\implies \color{red}{\ln(2+e^y)}=-\left(\frac{x}{e^x}+\frac{1}{e^x}\right)+C\\ &\implies y=\ln\left(\exp\left(-\left\{\frac{x}{e^x}+\frac{1}{e^x}\right\}+C\right )-2\right) \end{aligned}$

The problem is that if you let $\varphi\mapsto 2+e^y$ in your integral you get $d\varphi=e^y~dy$ so you should have $\int\frac{e^y}{2+e^y}dy\stackrel{\varphi=2+e^y}{\l ongmapsto}\int\frac{d\varphi}{\varphi}$

My b and b.

There's not really problem.

The result given by danny is correct. What you've done (which is fine by the way) is to make explicit how the integration is done.

**geton** · December 25th 2008, 10:33 PM

Originally Posted by mr fantastic

No.

First you note that $4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t$ .

Then you make the substitution ${\color{red}u = \sin t}$ (NOT $u = \sin t$ which was no doubt a typo by danny). Note that $dt = \frac{du}{\cos t}$ .

Thanks.

**Mathstud28** · December 25th 2008, 10:37 PM

Originally Posted by mr fantastic

My b and b.

There's not really problem.

The result given by danny is correct. What you've done (which is fine by the way) is to make explicit how the integration is done.

Yes. I am sorry. I missed Danny's post.

**Danny** · December 26th 2008, 06:57 AM

Originally Posted by mr fantastic

No.

First you note that $4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t$ .

Then you make the substitution ${\color{red}u = \sin t}$ (NOT $u = \cos t$ which was no doubt a typo by danny). Note that $dt = \frac{du}{\cos t}$ .

Yep - a typo on my part.

It should be $u = \sin t$

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