Question

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

Answer 1

thats a little section beside the sin cos stuff eh

Answer 2

since sin and cos = each other at pi/4; take your integrals from 0 to pi/4

Answer 3

[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]

Answer 4

to revolve it around the x axis; we do a sum of areas

[S] 2pi [f(x)]^2 dx

Answer 5

take the cos first and subtract out the sin next; like cutting a hole out of a donut

Answer 6

ack!!... pi [f(x)]^2 lol

Answer 7

pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]

Answer 8

makes sense?

Answer 9

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 \[\int\limits_{0}^{\pi/2}\int\limits_{\sin x}^{\cos x} dy dx\]

Answer 10

seems to me because the area enclosed by y-axis the interval should be from pi/4 yo pi/2...

Answer 11

nice drawing ! so, shaded area - [pi/4,pi/2]. am I wrong?

Answer 12

amistre, is that a (cos(x))² or is it cos (x²)

Answer 13

your wrong :) 0 to pi/4

Answer 14

cos^2(x)

Answer 15

ok

Answer 16

i'm sorry, it's from o to pi/4 :)

Answer 17

the double integral? dunno bout it ;)

Answer 18

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.

Answer 19

cos(2t) = 2cos^2 - 1

cos^2 = (1+cos(2t))/2 might help

Answer 20

just havent had any experience with doubles yet ;)

Answer 21

for part A did you come up with sqrt(2) - 1?

Answer 22

[S] cos^2(x) dx

[S] 1/2 dt + [S] 2cos(2x)/4 dt ; [0,pi/4] is the first one right?

Answer 23

whaa !?!

Answer 24

[S] cos(t) dt -> sin(t). sin(pi/4) - sin(0) = 1/sqrt(2) - 0

Answer 25

or .414

Answer 26

[S] sin(t) dt -> -cos(t). -cos(pi/4) + cos(0) = -1/sqrt(2) +1

Answer 27

i got sqrt (2) - 1 for question a

Answer 28

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?

Answer 29

1/sqrt(2) + 1/sqrt(2) -1
=2/sqrt (2) - 1
=sqrt (2) - 1

Answer 30

i see it ;)

Answer 31

melor got it right :)

Answer 32

okay I understand part a :)

Answer 33

i invented a sqrt(2) in the numerator lol

Answer 34

part b is integrateing the areas of the "donut", ie torus

Answer 35

pi [S] cos^2 - sin^2 from 0 to pi/4 right?

Answer 36

what's the difference between that, and having a 2 pi in front? I always get mixed up when to put pi or 2pi

Answer 37

which is just cos(2t) thich integrates to (pi/2).sin(2t) from 0 to pi/4

Answer 38

2pi is a surface area

Answer 39

you want to add areas to get volume.... area of a circle = pi [f(x)]^2

Answer 40

okay!

Answer 41

2pi [f(x)] amounts to surface area :)

Answer 42

adding together all the circumfrences of the circles :)

Answer 43

so we're working with pi[S] cos²x - sin²x dx [0,pi/4] ?

Answer 44

yes

Answer 45

gosh, I feel silly. How do you integrate that? :/

Answer 46

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

Answer 47

keep the top 2 to use in the integral and slide out the bottom to under the pi

Answer 48

we get in the end; (pi/2) sin(2x) right?

Answer 49

go you recall your trig identitites?

Answer 50

ehh, cos²x-sin²x= cos(2x)?

Answer 51

the volume should be

pi
------
2sqrt(2)

Answer 52

cos(x+y)=cos x cos y - sin x sin y
substitue x into y you get,
cos (2x) = cos^2 - sin^2 :)

Answer 53

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)

Answer 54

D(sin(2x)) = 2 cos(2x)

Answer 55

2 cos(2x)
-------- = cos(2x) right?
2

Answer 56

I plugged pi[S] cos²x - sin²x dx [0,pi/4] into wolframalpha and it gave me pi/2?

Answer 57

pi/2 [S] 2cos(2x) dx -> pi/2 sin(2x)

Answer 58

at 0 = its = 0

Answer 59

so the volume of the solid is pi/2 * sin(2(45)) = pi/2

Answer 60

sin(2(pi/4)) = sin(pi/2) = 1

Answer 61

so yes; the volume of the solid of rotation is pi/2 :)

Answer 62

sweet!

Answer 63

i gots no idea what c is saying

Answer 64

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 :(

Answer 65

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