Question

find the volume of the solid bounded by z=-y, the xy plane, and r=4cos(theta)

Answer 1

We need a function, it seems best to use polar coordinates no?

Answer 2

yes

Answer 3

So, convert z = -y to
\[\huge z = -r\sin \theta\]

Answer 4

and xy plane mean lower limit of z =0

Answer 5

Wait, no.

Answer 6

The function to integrate is z...
\[\huge z = f(r,\theta) = -4\sin \theta\]

Answer 7

Wait... stop... we'll redo this.

Answer 8

(sorry, I actually just woke up :) )

Answer 9

Let's do it in rectangular instead...

Answer 10

And convert \[\large r = 4 \cos \theta\]into rectangular.

Answer 11

Is it even possible?

Answer 12

you mean we convet it into rectangular respect to x,y,z?

Answer 13

Yeah...

Answer 14

or cylindrical r, theta and z?

Answer 15

wait, let me have a good think...

Answer 16

ok, take your time, I appreciate your help

Answer 17

Okay, definitely better to use rectangular.
But first, I need you (not really, lol, but do it anyway :) )
to convert
\[\huge r = 4\cos \theta\]
into rectangular form.

Answer 18

x = r costheta= 4cos^2 theta?

Answer 19

yeah... using those, and also
\[\huge r = \sqrt{x^2 + y^2}\]

Answer 20

y =4cos sin

Answer 21

uhh... be a little more clever :P
\[\huge \cos \theta = \frac{x}{r}\]

Answer 22

don't get, hihihi the purpose of this step

Answer 23

Well...
\[\huge r = 4\left(\frac{x}{r}\right)\]
\[\huge r^2 = 4x\]
\[\huge x^2 + y^2 = 4x\]

Answer 24

oh, to get z = x^2 +y^2 -4x?

Answer 25

No... this is a region,
\[\huge x^2 - 4x + y^2 = 0\]

Answer 26

\[\huge (x-2)^2 + y^2=4\]

Answer 27

yes, but it is a upper limit of z, too?

Answer 28

This is just the region over which we integrate, the real function that we're integrating is
\[\huge z = f(x,y) = -y\]

Answer 29

ok, you have a circle there

Answer 30

yes.

Answer 31

\[\huge \iint\limits_R -y \ dA\]

Answer 32

so the function to take integral yes, got it ,

Answer 33

Where R is the region bounded by that circle.

Answer 34

tripple or double?

Answer 35

Just double. We're only taking a volume.

Answer 36

what different between double and tripple, when we use this or that? both them compute volume ?

Answer 37

Triple Integral actually computes Hyper-volume, but may be used to compute volume. We can solve this using a triple integral, but it will lead to this double integral anyway.

Answer 38

wow, just get from you

Answer 39

hmm?

Answer 40

we just have the function, how to set the limit?

Answer 41

We have the region... R the circle defined by
\[\huge (x-2)^2+y^2 = 4\]

Answer 42

do we have to convert back to r and theta?

Answer 43

No. You can forget about those, actually :)

Answer 44

why? it's easier, right? we have circle r from 0 to2, theta from 0 to 2pi and the function is -y means -rsin theta, is it not enough to compute?

Answer 45

Yes, unfortunately, the usefulness of using polar coordinates diminishes rapidly when your circle is not centred around the pole (origin)

Answer 46

really? oh yea, you are right!!! :)

Answer 47

I told you... it's Terence :D
Now, the region has y running from \(\large -\sqrt{4-(x-2)^2}\)
to \(\large \sqrt{4-(x-2)^2}\)

Answer 48

hihihi.... yes, Terence!!!

Answer 49

And x runs from 0 to 4

Answer 50

Because the circle is centred around the point (2,0)

Answer 51

got it.

Answer 52

so the whole thing is int_from 0 to 4 int_ (the bunch you give out above) of -y dy dx

Answer 53

\[\huge \int_{0}^4\int_{-\sqrt{4-(x-2)^2}}^{\sqrt{4-(x-2)^2}}-y \ dy dx\]

Answer 54

Might look hard, but this integral should be easy...

Answer 55

hi Terence, how can you have that keen flexibility, how can you figure out right after look at the problem and decide which method should be use, teeeaaach me , please

Answer 56

A diet of noodles? I really don't know... it sort of just happens :)
I'd say experience, but I don't have that much.

I do stand by that epithet, though... experience is the best teacher :D

Answer 57

Lol, got it, thanks a lot Terence,

Answer 58

No problem :) I'm going to sign off for now.
---------------------------------------------
Terence out