Question

Having difficulty finding good limits of integration for finding a volume using a triple integral. f(x, y, z) = 1. Bounded by x^2 + y^2 = 9; z = 1; and x+z=5. So far, I have -3<=y<=3; 5-x<=z<=1; and -sqrt(9-y^2)<=x<=sqrt(9-y^2).

Problem is that I can't get to the final integration without ending up with a really nasty nasty integral involving the sqrt(9-y^2).

I'm trying dz, dz, dy so that'd be type 1.

Not necessarily looking for a solution, just trying to find better limits of integration or maybe I'm looking at this wrong.

Answer 1

\(\sqrt{9 - y^{2}}\) is NOT a nasty integral. You didn't expect a result with no \(\pi\) in it, did you? You'll need some trig functions somehow. Try substituting \(y=3\sin(\theta)\), or something like that.

Answer 2

so we have a cylindar sliced by a plane

Answer 3

Yea, I can see it in my head. I can see what's going on there. My integration skills are not the best, I'll definitely give you that. I wanted to try to turn it into a cylindrical integral there, but not sure where to go with that.

Answer 4

your choice of dx,dy,dz will affect the integration ... since 5 is above 1, and greater than 3.

Answer 5

its the top part isnt it ... or the bottom part

Answer 6

its the top part, liek a crystal award

Answer 7

z moves from 1 to x+5, or -x+5 but thats immaterial in this regards

y moves from -sqrt(9-x^2) to +sqrt(9-x^2)

x moves from -3 to 3

Answer 8

ok, we have the same movements

Answer 9

The attached shows where I get to with this. I'd like to turn the sqrt into something involving r and theta, but not sure how to go about that.

Answer 10

\[\int_V dz~dy~dx\]
\[\int_{y=-3}^{y=3}\int_{x=-\sqrt{9-y^2}}^{x=\sqrt{9-xy^2}}\int_{z=1}^{z=-x+5} dz~dy~dx\]

\[\int_{x=-3}^{x=3}\int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}}-x+4~dy~dx\]
\[\int_{x=-3}^{x=3}\left[-\frac12 y^2+4y\right]_{\pm \sqrt{9-y^2}}~dx\]

Answer 11

8 sqrt(9-y^2)

its not that bad lol

Answer 12

In your middle step there, I'm confused where the y^2 term is coming from. If integrating -x+4 WRT y, wouldn't that give -xy + 4y?

Answer 13

if a = -b

f(b) - f(-b)

(b)^2 - (-b)^2 = 0

Answer 14

yeah, i cant fight this lag and try to keep track of things ...

Answer 15

dz dx dy

Answer 16

4-x dx is 4x -1/2 x^2

Answer 17

1 dz, = z, 5-x -1 = 4-x

4 - x dx = 4x - 1/2 x2

8sqrt(9-y^2) dy

Answer 18

as tkhunny pointed out, its a simple trig sub if need be

Answer 19

since z is a function of x, it just makes sense to have dx after dz ... letting x be a function of y

Answer 20

we could polar it: since dx dy = r dr dt and dz = dz ... if thats what you want to go with

Answer 21

since x=rcos(t)

z = 1 to 5-r cos(t)
t = 0 to 2pi
r = 0 to 3

\[\int_{r=0}^{r=3}\int_{t=0}^{t=2\pi}\int_{z=1}^{z=5-rcos(t)}r~dz~dt~dr\]
\[\int_{r=0}^{r=3}\int_{t=0}^{t=2\pi}r~\int_{z=1}^{z=5-rcos(t)}~dz~dt~dr\]
\[\int_{r=0}^{r=3}\int_{t=0}^{t=2\pi}r~(4-rcos(t))~dt~dr\]
\[\int_{r=0}^{r=3}8\pi~r~dr\]

Answer 22

That makes infinitely more sense. Don't know why I didn't think about it like that. I was going to take the inner integral and then just turn it into a polar double integral which I think is ultimately what's happening.

Answer 23

cylindar coordinant system