Question

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.

Answer 1

Two factors are possible.

One is that no one with this level of expertise has been online to see your question.

The other possibility is that those who might help you are waiting to see what work you've done to solve these problems.

Answer 2

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

Answer 3

See also reply given by Damon below.

Answer 4

Please submit all complaints in triplicate using heavy carbon paper and obtain signatures from all subordinates, notarized. :(

Answer 5

Carbon paper? What's that? LOL!

Your answer is great, Damon! :-)

Answer 6

you put it in a typewriter

Answer 7

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