Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line y = 3 sin x, y = 3 cos x, 0 ≤ x ≤ π/4; about y = −1

Answer 1

Here's the graph

Answer 2

I know I have to use washer method by using A(x) = pi(rout)^2 - pi(rin)^2 and integrate that from 0 to pi/4 to get the volume

Answer 3

Hmm yah the washer method seems like the way to go :) since both functions were given in terms of x. Hmmm let's see..

Answer 4

so the outer radius would be.. 3cosx+1 and the inner radius would be 3sinx+1.. right ?

Answer 5

Yessss, good good c:

Answer 6

so you'd go ahead and square and have the equation pi(9cos^2x + 6cosx +1) - pi(9sin^2x + 6sinx + 1) for your area of the washer right ?

Answer 7

I'd probably setup the integral before squaring, but ya that seems fine.

Answer 8

Okaaay so then we'd have.. \[\Pi \int\limits_{0}^{\Pi/4}(9\cos ^{2}x + 6cosx +1) - (9\sin ^{2}x + 6sinx +1) dx\]
You can pull the pi out front, right ?

Answer 9

factoring it out of each term? ya looks good!

Answer 10

Im used totheformula naturally having pi out front xD But yeah, Im so used to picking some wrong radius or mixing something up with these, so I like to watch xD

Answer 11

There is a different pi button that looks better :) lol
If you're typing it use pi with lowercase P, gives you \(\large \pi\)

Answer 12

Okay and then i can separate it to two integrals right ?
\[\pi \int\limits_{0}^{\pi/4}(9\cos ^{2}x + 6cosx +1)dx - \pi \int\limits_{0}^{\pi/4}(9\sin ^{2}x + 6sinx + 1)dx\]
Yes?

Answer 13

You can, but I don't think we want to do that.

I think we want to rewrite our sin^2x using our `Square Identity`.
Then we can have just cos^2x's and deal with all the squares at the same time.

Answer 14

\[\Large \sin^2x=1-\cos^2x\]Right? :x

Answer 15

Oh yeah, okay.

Answer 16

so then we'd have
\[\pi \int\limits_{0}^{\pi/4}(9\cos ^{2}x + 6 cosx +1) - ((9-9\cos ^{2}x) + 6sinx +1) dx\]

Answer 17

Yes ?

Answer 18

yes c: simplify it down a bit! heh

Answer 19

So the ones cancel and we'll have..
\[\pi \int\limits_{0}^{\pi/4}(18\cos ^{2}x + 6cosx - 6sinx - 9)dx\]
right?

Answer 20

good good.
From there, do you remember your `Half-Angle Identity`?

Answer 21

Power-reducing?

Answer 22

yes

Answer 23

Lol, yeah, thought you meant that :P

Answer 24

It's called Half-Angle silly D:

Answer 25

cos^2x = 1/2(1 + cos(2a)) right ?

Answer 26

yah c:

Answer 27

There ya go :P And thats soooo power-reducing D:

Answer 28

I guess I think of a different formula when I think of `Power Reducing`.
When you have to solve something like:\[\Large \int\limits \cos^8x\;dx\]It involves using a formula to reduce the power a bunch of times :D

Answer 29

\[\cos ^{2}x=\frac{ 1+\cos2x }{ 2 }\]power-reduce
\[\cos \frac{ x }{ 2 }=\pm \sqrt{\frac{ 1+cosx }{ 2 }}\]half-angle :D
Lol, but don't mind me, im just being weird o.o

Answer 30

They're the same formula!! :O Square the second one :3 what do you get?
And see how the power is doubled in both of them?

pshhhhh

Answer 31

not the power, i mean the angle is doubled.. blah

Answer 32

lol

Answer 33

so the 18 will become a 9, the 9's cancel and we have
\[\pi \int\limits_{0}^{\pi/4}(9\cos(2x) + 6cosx -6sinx)dx\]
correct ?

Answer 34

mmmm yah that looks good.

Answer 35

Smooth sailing from there :3

Answer 36

These types of problems can be rather annoying.
Since they're so long, it's very easy to make a mistake along the way.

Do you have an answer key by chance? So we can check our answer at the end.

Answer 37

I usually just make a mistake with choosing the correct radii x_x

Answer 38

I do not, its on webassign. But I can submit it to check it once I get there.

Answer 39

So the answer is 9(pi)/2 - 6(pi)(sqrt2) ?

Answer 40

Let's check.

Answer 41

Whoops, hang on xD

Answer 42

\[\Large \pi\left[\left(\frac{9}{2}+6\frac{\sqrt2}{2}+6\frac{\sqrt2}{2}\right)-\left(\color{#F35633}{6}\right)\right]\]

Answer 43

When you plug in your limits, you should get something like that.
Oops I simplified down the first term. but whatever.

The pink value is our lower limit.

Answer 44

Our lower limit.. I thought it was from 0 to pi/4 ?

Answer 45

\(\large \cos0=?\)

Answer 46

gahhhhh, right, its 1

Answer 47

\[\pi(\frac{ 9 \sin2x}{ 2 }+6sinx+6cosx)\]

Answer 48

Crap, my 2nd part got deleted.

Answer 49

aw :c

Answer 50

\[\pi(\frac{ 9 }{ 2 }+\frac{ 12\sqrt{2} }{ 2 }-\frac{ 12 }{ 2 })\]

Answer 51

>.>

Answer 52

so 9(pi)/2 + 6(pi)sqrt 2) - 6(pi)

Answer 53

Pretty much. Whatever simplified form you like most xD

Answer 54

bah thats so ugly :3 lol

Answer 55

Ikr? D:

Answer 56

But it's right. Yay ! Thanks guys. (:

Answer 57

I simplified it down to this, I think it looks a tad nicer +_+\[\Large \frac{\pi}{2}\left[12\sqrt2-3\right]\]

Answer 58

Oh good job \c:/

Answer 59

Okay, how id have left it doesnt look as neat as that xD