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anonymous
 5 years ago
Let R be the shaded region in the first quadrant enclosed by the yaxis 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 xaxis
c)Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the xaxis are squares
anonymous
 5 years ago
Let R be the shaded region in the first quadrant enclosed by the yaxis 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 xaxis c)Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the xaxis are squares

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2thats a little section beside the sin cos stuff eh

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2since sin and cos = each other at pi/4; take your integrals from 0 to pi/4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2[S] cos(t) dt  [S] sin(t) dt ;[0,pi/4]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2to revolve it around the x axis; we do a sum of areas [S] 2pi [f(x)]^2 dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2take the cos first and subtract out the sin next; like cutting a hole out of a donut

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2ack!!... pi [f(x)]^2 lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2pi [S] cos(x)^2 dx  [S] sin(x)^2 dx ; [0,pi/4]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0seems to me because the area enclosed by yaxis the interval should be from pi/4 yo pi/2...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nice drawing ! so, shaded area  [pi/4,pi/2]. am I wrong?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre, is that a (cos(x))² or is it cos (x²)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2your wrong :) 0 to pi/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm sorry, it's from o to pi/4 :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2the double integral? dunno bout it ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought the question want to integrate between cos x , sin x and xaxis. silly me :P you can find area under curve using double integration.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2cos(2t) = 2cos^2  1 cos^2 = (1+cos(2t))/2 might help

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2just havent had any experience with doubles yet ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for part A did you come up with sqrt(2)  1?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2[S] cos^2(x) dx [S] 1/2 dt + [S] 2cos(2x)/4 dt ; [0,pi/4] is the first one right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2[S] cos(t) dt > sin(t). sin(pi/4)  sin(0) = 1/sqrt(2)  0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2[S] sin(t) dt > cos(t). cos(pi/4) + cos(0) = 1/sqrt(2) +1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got sqrt (2)  1 for question a

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.21/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
 5 years ago
Best ResponseYou've already chosen the best response.01/sqrt(2) + 1/sqrt(2) 1 =2/sqrt (2)  1 =sqrt (2)  1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2melor got it right :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay I understand part a :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2i invented a sqrt(2) in the numerator lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2part b is integrateing the areas of the "donut", ie torus

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2pi [S] cos^2  sin^2 from 0 to pi/4 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what's the difference between that, and having a 2 pi in front? I always get mixed up when to put pi or 2pi

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2which is just cos(2t) thich integrates to (pi/2).sin(2t) from 0 to pi/4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.22pi is a surface area

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2you want to add areas to get volume.... area of a circle = pi [f(x)]^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.22pi [f(x)] amounts to surface area :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2adding together all the circumfrences of the circles :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we're working with pi[S] cos²x  sin²x dx [0,pi/4] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gosh, I feel silly. How do you integrate that? :/

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2cos^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
 5 years ago
Best ResponseYou've already chosen the best response.2keep the top 2 to use in the integral and slide out the bottom to under the pi

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2we get in the end; (pi/2) sin(2x) right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2go you recall your trig identitites?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ehh, cos²xsin²x= cos(2x)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2the volume should be pi  2sqrt(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(x+y)=cos x cos y  sin x sin y substitue x into y you get, cos (2x) = cos^2  sin^2 :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2cos(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
 5 years ago
Best ResponseYou've already chosen the best response.2D(sin(2x)) = 2 cos(2x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.22 cos(2x)  = cos(2x) right? 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I plugged pi[S] cos²x  sin²x dx [0,pi/4] into wolframalpha and it gave me pi/2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2pi/2 [S] 2cos(2x) dx > pi/2 sin(2x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2so the volume of the solid is pi/2 * sin(2(45)) = pi/2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2sin(2(pi/4)) = sin(pi/2) = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2so yes; the volume of the solid of rotation is pi/2 :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2i gots no idea what c is saying

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
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