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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.
[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
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.
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.