anonymous
  • anonymous
Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y=sinx and y=cosx. a) Find the area of R b)Find the volume of the solid generated when R is revolved about the x-axis c)Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
thats a little section beside the sin cos stuff eh
amistre64
  • amistre64
since sin and cos = each other at pi/4; take your integrals from 0 to pi/4
amistre64
  • amistre64
[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
to revolve it around the x axis; we do a sum of areas [S] 2pi [f(x)]^2 dx
amistre64
  • amistre64
take the cos first and subtract out the sin next; like cutting a hole out of a donut
amistre64
  • amistre64
ack!!... pi [f(x)]^2 lol
amistre64
  • amistre64
pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]
amistre64
  • amistre64
makes sense?
anonymous
  • anonymous
i think the integral will involve integration from x =0 to 90 degree since it's given that the problem is about to solve at the first quadrant. for question a) integration \[\int\limits_{0}^{\pi/2}\int\limits_{\sin x}^{\cos x} dy dx\]
anonymous
  • anonymous
seems to me because the area enclosed by y-axis the interval should be from pi/4 yo pi/2...
amistre64
  • amistre64
1 Attachment
anonymous
  • anonymous
nice drawing ! so, shaded area - [pi/4,pi/2]. am I wrong?
anonymous
  • anonymous
amistre, is that a (cos(x))² or is it cos (x²)
amistre64
  • amistre64
your wrong :) 0 to pi/4
amistre64
  • amistre64
cos^2(x)
anonymous
  • anonymous
ok
anonymous
  • anonymous
i'm sorry, it's from o to pi/4 :)
amistre64
  • amistre64
the double integral? dunno bout it ;)
anonymous
  • anonymous
i thought the question want to integrate between cos x , sin x and x-axis. silly me :P you can find area under curve using double integration.
amistre64
  • amistre64
cos(2t) = 2cos^2 - 1 cos^2 = (1+cos(2t))/2 might help
amistre64
  • amistre64
just havent had any experience with doubles yet ;)
anonymous
  • anonymous
for part A did you come up with sqrt(2) - 1?
amistre64
  • amistre64
[S] cos^2(x) dx [S] 1/2 dt + [S] 2cos(2x)/4 dt ; [0,pi/4] is the first one right?
amistre64
  • amistre64
whaa !?!
amistre64
  • amistre64
[S] cos(t) dt -> sin(t). sin(pi/4) - sin(0) = 1/sqrt(2) - 0
anonymous
  • anonymous
or .414
amistre64
  • amistre64
[S] sin(t) dt -> -cos(t). -cos(pi/4) + cos(0) = -1/sqrt(2) +1
anonymous
  • anonymous
i got sqrt (2) - 1 for question a
amistre64
  • amistre64
1/sqrt(2) - (-1/sqrt(2) +1) 1/sqrt(2) + 1/sqrt(2) -1 (2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1 right?
anonymous
  • anonymous
1/sqrt(2) + 1/sqrt(2) -1 =2/sqrt (2) - 1 =sqrt (2) - 1
amistre64
  • amistre64
i see it ;)
amistre64
  • amistre64
melor got it right :)
anonymous
  • anonymous
okay I understand part a :)
amistre64
  • amistre64
i invented a sqrt(2) in the numerator lol
amistre64
  • amistre64
part b is integrateing the areas of the "donut", ie torus
amistre64
  • amistre64
pi [S] cos^2 - sin^2 from 0 to pi/4 right?
anonymous
  • anonymous
what's the difference between that, and having a 2 pi in front? I always get mixed up when to put pi or 2pi
amistre64
  • amistre64
which is just cos(2t) thich integrates to (pi/2).sin(2t) from 0 to pi/4
amistre64
  • amistre64
2pi is a surface area
amistre64
  • amistre64
you want to add areas to get volume.... area of a circle = pi [f(x)]^2
anonymous
  • anonymous
okay!
amistre64
  • amistre64
2pi [f(x)] amounts to surface area :)
amistre64
  • amistre64
adding together all the circumfrences of the circles :)
anonymous
  • anonymous
so we're working with pi[S] cos²x - sin²x dx [0,pi/4] ?
amistre64
  • amistre64
yes
anonymous
  • anonymous
gosh, I feel silly. How do you integrate that? :/
amistre64
  • amistre64
cos^2 - sin^2 = cos(2x) right? 2 cos(2x) ups to sin(2x) so we need to multiply by a comvenient form of 1; say 2/2
amistre64
  • amistre64
keep the top 2 to use in the integral and slide out the bottom to under the pi
amistre64
  • amistre64
we get in the end; (pi/2) sin(2x) right?
amistre64
  • amistre64
go you recall your trig identitites?
anonymous
  • anonymous
ehh, cos²x-sin²x= cos(2x)?
amistre64
  • amistre64
the volume should be pi ------ 2sqrt(2)
anonymous
  • anonymous
cos(x+y)=cos x cos y - sin x sin y substitue x into y you get, cos (2x) = cos^2 - sin^2 :)
amistre64
  • amistre64
cos(a+b) = cos(a)cos(b) - sin(a)sin(b) cos(a+a); or cos(2a) = cos(a)cos(a)-sin(a)sin(a cos(2a)=cos^2(a)-sin^2(a)
amistre64
  • amistre64
D(sin(2x)) = 2 cos(2x)
amistre64
  • amistre64
2 cos(2x) -------- = cos(2x) right? 2
anonymous
  • anonymous
I plugged pi[S] cos²x - sin²x dx [0,pi/4] into wolframalpha and it gave me pi/2?
amistre64
  • amistre64
pi/2 [S] 2cos(2x) dx -> pi/2 sin(2x)
amistre64
  • amistre64
at 0 = its = 0
amistre64
  • amistre64
so the volume of the solid is pi/2 * sin(2(45)) = pi/2
amistre64
  • amistre64
sin(2(pi/4)) = sin(pi/2) = 1
amistre64
  • amistre64
so yes; the volume of the solid of rotation is pi/2 :)
anonymous
  • anonymous
sweet!
amistre64
  • amistre64
i gots no idea what c is saying
anonymous
  • anonymous
c is asking to take the graph that we had, using the area R as the base of a solid, placing square cross sections on top of the graph to create a solid... haha I have no clue how to explain it. and I don't know how to solve it :(
anonymous
  • anonymous
I guess for a square, every side is equal, so the length of each individual square placed is going to fit depending on the base R

Looking for something else?

Not the answer you are looking for? Search for more explanations.