Surface area of cylinder x² + z² = 1 inside cylinder x² + y² = 2ay

[symmetric]

1. Homework Statement :

Find surface area of part of cylinder $x^2 + z^2 = 1$ that is inside the cylinder $x^2 + y^2 = 2ay$ and also in the positive octant ( $x \geq 0, y \geq 0, z \geq 0$ ). Assume a > 0.

Homework Equations

$$x^2 + z^2 = 1$$
$$x^2 + y^2 = 2ay$$
( $x \geq 0, y \geq 0, z \geq 0$ )
a > 0

The Attempt at a Solution

Cylinder expression $x^2 + z^2 = 1$ can be written as $z = \sqrt{ ( a^2 - x^2 ) }$.This surface can be projected on X-Y plane by using following parametrization -

x = x
y = y
z = f( x, y ) = $\sqrt{ a^2 - x^2 }$

Hence the surface area of cylinder $x^2 + z^2 = 1$ for above parametrization is given by integral -

Area integral = $\iint{\sqrt{ ( 1 + f'_{x})^2 + ( f'_{y})^2 }} dx dy$

Now, $\sqrt{ 1 + (f'_{x})^2 + ( f'_{y})^2 }$ = $\sqrt{ 1 + \frac{x^2}{a^2 - x^2} + 0 }$ = $\frac{a}{ \sqrt{ a^2 - x^2 }}$


Area integral = $\iint{ \frac{a}{ \sqrt{ a^2 - x^2 }}} dx dy$

The limits of this integral will be decided by second insecting cylinder $x^2 + y^2 = 2ay$

Working out for finding limits -

$y^2 + (-2a) y + x^2 = 0$ treating second order equation in ' y ' we can write -

$y = a \pm \sqrt { a^2 - x^2 }$ selecting positive root for given conditions for positive octant.

Hence limits are – for x -> 0 to a and for y -> 0 to $a + \sqrt { a^2 - x^2 }$

The integral becomes


Area integral = $\int_{x = 0}^{ x = a } \int_{ y = 0 }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx$

= $\int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } )}{ \sqrt { a^2 - x^2 }} dx$

= $a^2 \int_{0}^{a} \frac{ dx }{ \sqrt { a^2 - x^2 } } + a \int_{0}^{a}} dx$
= $a^2 \left[ \arcsin \left ( \frac{x}{a} \right) \right]_{0}^{a} + a^2$

Required area = $a^2 \left[ \frac{\pi}{2} +1 \right]$


The answer given in the book is $2a^2$ . Am I going wrong somewhere ? Please help.

 

[lanedance]

1. Homework Statement :

Find surface area of part of cylinder $x^2 + z^2 = 1$ that is inside the cylinder $x^2 + y^2 = 2ay$ and also in the positive octant ( $x \geq 0, y \geq 0, z \geq 0$ ). Assume a > 0.

Homework Equations

$$x^2 + z^2 = 1$$
$$x^2 + y^2 = 2ay$$
( $x \geq 0, y \geq 0, z \geq 0$ )
a > 0

The Attempt at a Solution

Cylinder expression $x^2 + z^2 = 1$ can be written as $z = \sqrt{ ( a^2 - x^2 ) }$.This surface can be projected on X-Y plane by using following parametrization -

x = x
y = y
z = f( x, y ) = $\sqrt{ a^2 - x^2 }$

shouldn't
$$z = f( x, y ) = \sqrt{ 1 - x^2 }$$?

also you need to assume a < 1, which you're doing, but should be aware of

otherwise i think your mothod is ok ;)

 

[symmetric]

@lanedance - Thanks for your reply.

You have pointed out correct typing mistake due to some 'copy paste'. Sorry for that.

Corrected problem is as follows -


1. Homework Statement :

Find surface area of part of cylinder $x^2 + z^2 = a^2$ that is inside the cylinder $x^2 + y^2 = 2ay$ and also in the positive octant ( $x \geq 0, y \geq 0, z \geq 0$ ). Assume a > 0.

Homework Equations

$$x^2 + z^2 = a^2$$
$$x^2 + y^2 = 2ay$$
( $x \geq 0, y \geq 0, z \geq 0$ )
a > 0

The Attempt at a Solution

Cylinder expression $x^2 + z^2 = a^2$ can be written as $z = \sqrt{ ( a^2 - x^2 ) }$.This surface can be projected on X-Y plane by using following parametrization -

x = x
y = y
z = f( x, y ) = $\sqrt{ a^2 - x^2 }$

Hence the surface area of cylinder $x^2 + z^2 = a^2$ for above parametrization is given by integral -

Area integral = $\iint{\sqrt{ ( 1 + f'_{x})^2 + ( f'_{y})^2 }} dx dy$

Now, $\sqrt{ 1 + (f'_{x})^2 + ( f'_{y})^2 }$ = $\sqrt{ 1 + \frac{x^2}{a^2 - x^2} + 0 }$ = $\frac{a}{ \sqrt{ a^2 - x^2 }}$


Area integral = $\iint{ \frac{a}{ \sqrt{ a^2 - x^2 }}} dx dy$

The limits of this integral will be decided by second insecting cylinder $x^2 + y^2 = 2ay$

Working out for finding limits -

$y^2 + (-2a) y + x^2 = 0$ treating second order equation in ' y ' we can write -

$y = a \pm \sqrt { a^2 - x^2 }$ selecting positive root for given conditions for positive octant.

Hence limits are – for x -> 0 to a and for y -> 0 to $a + \sqrt { a^2 - x^2 }$

The integral becomes


Area integral = $\int_{x = 0}^{ x = a } \int_{ y = 0 }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx$

= $\int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } )}{ \sqrt { a^2 - x^2 }} dx$

= $a^2 \int_{0}^{a} \frac{ dx }{ \sqrt { a^2 - x^2 } } + a \int_{0}^{a} dx$
= $a^2 \left[ \arcsin \left ( \frac{x}{a} \right) \right]_{0}^{a} + a^2$

Required area = $a^2 \left[ \frac{\pi}{2} +1 \right]$


The answer given in the book is $2a^2$ . Am I going wrong somewhere ? Please help.

 

[lanedance]

ok so i had another look & i think your lower limit for the y integration step is incorrect, it should also be dependent on x.

it may help to change the integration order, do x then y.

 

[lanedance]

to help see it, the vertical cylinder given the bounds is defined by the circle
$$x^2 + (y-a)^2 = a^2$$
so radius a, centred on (0,a)

 

[symmetric]

@lanedance

You are right. After analyzing the problem again, posting the corrected area integral -


Area integral = $$\int_{x = 0}^{ x = a } \int_{ y = a - \sqrt { a^2 - x^2 } }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx$$


= $$\int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } - a + \sqrt { a^2 - x^2 }) }{ \sqrt { a^2 - x^2 }} dx$$


= $$2a \int_{0}^{a} dx$$

Required area = $$2a^2$$

Which is same as required !

Above problem solved in cylindrical parametrization is posted here -

http://www.cramster.com/answers-may-10/calculus/solved-but-answer-not-problem-statement-find-the-surface-area-of-that-part-of_854407.aspx

In addition to above two methods I have also tried above problem using Green's method given here -
http://mathworld.wolfram.com/SteinmetzSolid.html

( answer by Green's Theorem method is posted in my next post )


I just want confirm that, everything is solved correctly, and steps are appropriate for above three methods.

 

[symmetric]

Method III - Using Green's theorem

Given equation $x^2 + z^2 = a^2$ can be parametrized as follows -

x = x
z = $\sqrt{ a^2 - x^2 }$

According to result obtained from Green's theorem we can write -

Area = $\int{ y ds }$


where ds = $$\sqrt{ 1 + \left(\frac{dz}{dx} \right )^2 } dx$$


ds = $$\frac{a}{\sqrt{ a^2 - x^2 } } dx$$


Now, y can be calculated from cylinder equation $x^2 + y^2 = 2ay$ as

$y = a \pm \sqrt{ a^2 - x^2 }$

Here we have mapped required surface area on x-y plane. Mapped area region is divided into two regions for which y takes two different values calculated. Hence, required area is given by -

Area = $$\left( \int_{x = 0}^{ x = a }\frac{a( a + \sqrt{ a^2 - x^2 } ) dx } {\sqrt{ a^2 - x^2 }}\right) - \left( \int_{x = 0}^{ x = a }\frac{a( a - \sqrt{ a^2 - x^2 } ) dx } {\sqrt{ a^2 - x^2 }}\right)$$


Negative sign of second integral indicates opposite orientation of second part of surface.

Area = $2a \int_{0}^{a} dx$

Area = $2a^2$


Corrections and suggestions will be highly appreciated.