Area of one cylinder inside another intersecting cylinder

In summary, the task is to find the surface area of the portion of a cylinder x^2 + z^2 = a^2 that lies within the cylinder x^2 + y^2 = 2ay and in the positive octant. The equations for the cylinders and given conditions are provided. The solution involves parametrizing the surface and setting up an integral to calculate the area. A correction is made for a typing error in the parametrization. It is also noted that the solution is only valid for a < 1.
  • #1
symmetric
9
0
1. Homework Statement :

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

Homework Equations



[tex]x^2 + z^2 = 1[/tex]
[tex]x^2 + y^2 = 2ay[/tex]
( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] )
a > 0

The Attempt at a Solution



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

x = x
y = y
z = f( x, y ) = [itex]\sqrt{ a^2 - x^2 }[/itex]

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

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

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


Area integral = [itex] \iint{ \frac{a}{ \sqrt{ a^2 - x^2 }}} dx dy[/itex]

The limits of this integral will be decided by second insecting cylinder [itex]x^2 + y^2 = 2ay[/itex]

Working out for finding limits -

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

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

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

The integral becomes


Area integral = [itex] \int_{x = 0}^{ x = a } \int_{ y = 0 }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx[/itex]

= [itex] \int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } )}{ \sqrt { a^2 - x^2 }} dx[/itex]

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

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


The answer given in the book is [itex]2a^2[/itex] . Am I going wrong somewhere ? Please help.
 
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  • #2
symmetric said:
1. Homework Statement :

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

Homework Equations



[tex]x^2 + z^2 = 1[/tex]
[tex]x^2 + y^2 = 2ay[/tex]
( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] )
a > 0

The Attempt at a Solution



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

x = x
y = y
z = f( x, y ) = [itex]\sqrt{ a^2 - x^2 }[/itex]
shouldn't
[tex]z = f( x, y ) = \sqrt{ 1 - x^2 }[/tex]?

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

otherwise i think your mothod is ok ;)
 
  • #3
@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 [itex]x^2 + z^2 = a^2[/itex] that is inside the cylinder [itex]x^2 + y^2 = 2ay[/itex] and also in the positive octant ( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] ). Assume a > 0.

Homework Equations



[tex]x^2 + z^2 = a^2[/tex]
[tex]x^2 + y^2 = 2ay[/tex]
( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] )
a > 0

The Attempt at a Solution



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

x = x
y = y
z = f( x, y ) = [itex]\sqrt{ a^2 - x^2 }[/itex]

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

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

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


Area integral = [itex] \iint{ \frac{a}{ \sqrt{ a^2 - x^2 }}} dx dy[/itex]

The limits of this integral will be decided by second insecting cylinder [itex]x^2 + y^2 = 2ay[/itex]

Working out for finding limits -

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

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

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

The integral becomes


Area integral = [itex] \int_{x = 0}^{ x = a } \int_{ y = 0 }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx[/itex]

= [itex] \int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } )}{ \sqrt { a^2 - x^2 }} dx[/itex]

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

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


The answer given in the book is [itex]2a^2[/itex] . Am I going wrong somewhere ? Please help.
 
  • #4
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.
 
  • #5
to help see it, the vertical cylinder given the bounds is defined by the circle
[tex] x^2 + (y-a)^2 = a^2[/tex]
so radius a, centred on (0,a)
 
  • #6
@lanedance

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


Area integral = [tex] \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 [/tex]


= [tex] \int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } - a + \sqrt { a^2 - x^2 }) }{ \sqrt { a^2 - x^2 }} dx [/tex]


= [tex]2a \int_{0}^{a} dx [/tex]

Required area = [tex] 2a^2[/tex]

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.
 
Last edited by a moderator:
  • #7
Method III - Using Green's theorem

Given equation [itex] x^2 + z^2 = a^2 [/itex] can be parametrized as follows -

x = x
z = [itex] \sqrt{ a^2 - x^2 }[/itex]

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

Area = [itex] \int{ y ds }[/itex]


where ds = [tex] \sqrt{ 1 + \left(\frac{dz}{dx} \right )^2 } dx[/tex]


ds = [tex] \frac{a}{\sqrt{ a^2 - x^2 } } dx [/tex]


Now, y can be calculated from cylinder equation [itex] x^2 + y^2 = 2ay[/itex] as

[itex] y = a \pm \sqrt{ a^2 - x^2 }[/itex]

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 = [tex]\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)[/tex]


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

Area = [itex] 2a \int_{0}^{a} dx [/itex]

Area = [itex] 2a^2[/itex]


Corrections and suggestions will be highly appreciated.
 

What is the formula for finding the area of one cylinder inside another intersecting cylinder?

The formula for finding the area of one cylinder inside another intersecting cylinder is (2πr1h1) + (2πr2h2) + (πr12 - πr22), where r1 and r2 are the radii of the two cylinders and h1 and h2 are the heights of the cylinders.

How do you determine the radii and heights of the two intersecting cylinders?

The radii and heights of the two intersecting cylinders can be determined by measuring them with a ruler or by using the given dimensions if they are provided. In some cases, the dimensions may need to be calculated using other given information such as the volume or surface area.

What is the difference between the volume and surface area of a cylinder?

The volume of a cylinder is the measure of the space inside the cylinder, while the surface area is the measure of the total area of the outer surface of the cylinder. The volume is measured in cubic units, while the surface area is measured in square units.

Can the area of one cylinder inside another intersecting cylinder be negative?

No, the area of one cylinder inside another intersecting cylinder cannot be negative. The area is always a positive value, even if the two cylinders do not intersect or if one cylinder is completely inside the other.

What are some real-life applications of the concept of intersecting cylinders?

The concept of intersecting cylinders has many real-life applications, such as in plumbing where pipes intersect, in architecture where columns intersect with beams, and in engineering where pipes or cables intersect with structures. It is also used in the design of machinery and vehicles, such as in the engine of a car where pistons intersect with cylinders.

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