Area of one cylinder inside another intersecting cylinder

[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&#039;_{x})^2 + ( f&#039;_{y})^2 }} dx dy

Now, \sqrt{ 1 + (f&#039;_{x})^2 + ( f&#039;_{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.