The region R, is bounded by the graphs of x = 5/3 y and the curve C given by x = (1+y^2)^(1/2), and the x-axis.
a) Set up and evaluate an integral expression with respect to y that gives the area of R.
b) Curve C is part of the curve x^2 - y^2 = 1, Show that x^2 - y^2 = 1 can be written as the polar equation r^2 = 1/(cos^2theta - sin^2theta).
c) Use the polar equation given in part b) to set up an integral expression with respect to theta that represents the area of R.
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a) Integrate [(1+y^2)^(1/2) - 5/3 y ] dy from y = 0 to y = 3/4. You can substitute y = sinh(t) to make the maths a bit easier.
b) Looks trivial
c) Integrate r dr dtheta from r = 0 to r(theta) and theta = the desired interval =
Integral 1/2 r(heta)^2dtheta
See also reply given by Damon below.
Please submit all complaints in triplicate using heavy carbon paper and obtain signatures from all subordinates, notarized. :(
Carbon paper? What's that? LOL!
Your answer is great, Damon! :-)
you put it in a typewriter
for a) the integral should be from 0 to .75 and it is of (1+y^2)-5/3*y dy
b) since x=rcos# and y=rsin#
and x^2-y^2=1
whe can find that r^2cos^2#-r^2sin^2# =1
from there it is pretty obvious...
the problem is d)... i can't do it... HELP IS REQUIRED!!
ooo and the # is theta :D