# volume of revolutions

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• Jan 28th 2009, 05:56 PM
razorfever
volume of revolutions
find the volume of the solid generated by revolving about the y-axis the region bounded by x^2 + y^2 = a^2 , y = b>0 , x = 0
for b<=y<=a
(a spherical cap)

i assumed that y=b>0 meant to only count the graph above y=b
the answer i got was:

Pi * [ (a^4 -(a^6 / 3)) - ( (a^2)*b - (b^3 / 3) ) ]

sorry about all the brackets , i tried to make it as clean as possible

can someone check whether this is the correct answer and if not point me in the right direction
• Jan 28th 2009, 06:06 PM
Jhevon
Quote:

Originally Posted by razorfever
find the volume of the solid generated by revolving about the y-axis the region bounded by x^2 + y^2 = a^2 , y = b>0 , x = 0
for b<=y<=a
(a spherical cap)

i assumed that y=b>0 meant to only count the graph above y=b
the answer i got was:
Pi * [ (a^4 -(a^6 / 3)) - ( (a^2)*b - (b^3 / 3) ) ]

sorry about all the brackets , i tried to make it as clean as possible

can someone check whether this is the correct answer and if not point me in the right direction

so assuming you drew the graph correctly, you will notice that the shell method seems like the more straightforward way to go here.

the integral we must compute is:

$\displaystyle V = 2 \pi \int rh~dx$

$\displaystyle = 2 \pi \int_0^{\sqrt{a^2 - b^2}} x(\sqrt{a^2 - x^2} - b)~dx$

$\displaystyle = 2 \pi \int_0^{\sqrt{a^2 - b^2}} x \sqrt{a^2 - x^2}~dx - 2 \pi \int_0^{ \sqrt{a^2 - b^2}} bx~dx$

the second integral is easy, for the first, a substiutiton of $\displaystyle u = a^2 - x^2$ will work nicely. or even $\displaystyle u^2 = a^2 - x^2$ if you're up for it
• Jan 28th 2009, 08:37 PM
razorfever
how does the graph look like
can you describe it in words
or post a link to it or something
i really want to understand the question ... not just solve it
• Jan 28th 2009, 08:44 PM
Jhevon
Quote:

Originally Posted by razorfever
how does the graph look like
can you describe it in words
or post a link to it or something
i really want to understand the question ... not just solve it

here is the graph (see below). the shaded region is what you are rotating about the y-axis. i made the radius of the circle you see and the position of the horizontal line arbitrary, so don't mind the values that you see, just note what the general shape will look like.

now, do you remember how the shell method says to proceed?
• Jan 29th 2009, 12:50 PM
razorfever
by solving the integral

$\displaystyle = 2 \pi \int_0^{\sqrt{a^2 - b^2}} x \sqrt{a^2 - x^2}~dx - 2 \pi \int_0^{ \sqrt{a^2 - b^2}} bx~dx$

the answer i get is:

2Pi * ( (2/3)a^3 * (-sqrt(a^2-b^2) - 1))

is this the correct answer?
• Jan 31st 2009, 11:47 PM
mr fantastic
Quote:

Originally Posted by razorfever
find the volume of the solid generated by revolving about the y-axis the region bounded by x^2 + y^2 = a^2 , y = b>0 , x = 0
for b<=y<=a
(a spherical cap)

i assumed that y=b>0 meant to only count the graph above y=b
the answer i got was:

Pi * [ (a^4 -(a^6 / 3)) - ( (a^2)*b - (b^3 / 3) ) ]

sorry about all the brackets , i tried to make it as clean as possible

can someone check whether this is the correct answer and if not point me in the right direction

Duplicate post. (Headbang)

Things will be too messy trying to move the replies to here http://www.mathhelpforum.com/math-he...ut-y-axis.html and then edit so that it all makes sense.

Easier just to close this thread.