GR0568 #16

Forum for the GRE subject test in mathematics.
thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am

GR0568 #16

Postby thmsrhn » Fri Mar 26, 2010 8:00 am

Please help me solve the following problem.

"Whatr is the volume of the solid formed by revolving about the x axis the region in the first quadrant of the xy plane bounded by co ordinate axes and the graph of the equation y=(1+x^2)^(-.5)?

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

Re: GR0568 #16

Postby origin415 » Fri Mar 26, 2010 8:49 am

For volumes of solids of revolution, you want to create a new function corresponding to the cross-sectional area of the object, and integrate over that. Your function f(x) corresponds to the radius, so your integral will be

\pi \int (f(x))^2 dx

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

Re: GR0568 #16

Postby mathQ » Fri Mar 26, 2010 12:00 pm

Volume = Int pi * y^2 dx | (inf, -inf)


with that ans should be pi^2

thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am

Re: GR0568 #16

Postby thmsrhn » Fri Mar 26, 2010 2:27 pm

I know rite?!, but that ain t the answer. it ((pi)^2)/2!!

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: GR0568 #16

Postby joey » Fri Mar 26, 2010 5:29 pm

Only the part in the first quadrant is revolved. Use
A=\pi \int_0^\infty \left[f(x)\right] ^2dx.

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

Re: GR0568 #16

Postby mathQ » Sun Mar 28, 2010 1:21 am

Thanks joey.

the limit should be 0, Inf

with that ans is pi^2/2




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