Math Help - integral volume

1. integral volume

I want to find volume through integration.
Plane: $xy$
Paroloid: $z=9-x^2-y^2$

I tried to do task with double integrals, but ended up having 0 for an answer.

2. Originally Posted by Revy
I want to find volume through integration.
Plane: $xy$
Paroloid: $z=9-x^2-y^2$

I tried to do task with double integrals, but ended up having 0 for an answer.
first draw your self that volume (no need but can help a lot of times )

$z = 9-x^2-y^2$

so it look down and you need just to find volume of it from z= 9 to z= 0 am I correct ?

okay just use Cylindrical coordinates

your $r = 3$ so it means that goes from 0 to 3

and $\theta$ goes from 0 to $2\pi$

so you have

$\displaystyle \int _{0}^{2\pi } d\theta \int _0 ^{3} (9- r^2 ) r \;dr = \int _{0}^{2\pi } d\theta \big{\{} 9\codt \int _0 ^3 r \; dr - \int _0 ^3 r^3 \;dr \big{\}}=$

$\displaystyle \int _{0}^{2\pi } ( \frac {81}{2} - \frac {81}{4} ) \; d \theta = 2\cdot \pi \cdot \frac {81}{4} = \frac {81 \pi }{2}$

3. Originally Posted by yeKciM
first draw your self that volume (no need but can help a lot of times )
Well, I imagine how it looks like.

Originally Posted by yeKciM
you need just to find volume of it from z= 9 to z= 0 am I correct ?
Yes

Originally Posted by yeKciM
$\displaystyle \int _{0}^{2\pi } d\theta \int _0 ^{3} (9- r^2 ) r \;dr = . . .$
I still got 0 xD

4. i edited that up there to the end ...

5. Hello, Revy!

$\text{Volume between the }xy\text{-plane and the parabolid }z\:=\:9\:-x^2-y^2$

We have a parabolic "dome".
The base is the circle $x^2+y^2 \:=\:9$ on the $xy$-plane.
Its vertex is $(0,0,9).$

Consider its trace on the $xz$-plane.

Code:
      z
|
9 **
|:::*
|:::::*
|::::::*
|:::::::
- - + - - - * - x
|       3

In the first quadrant, we have: . $z \:=\:9 - x^2$

We revolve this region about the $z$-axis.

Cylindrical shells: . $\displaystyle V \;=\;2\pi \int^b_a x\!\cdot\!f\!(x)\,dx$

$\text{We have: }\;\displaystyle V \;=\;2\pi\int^3_0x\left(9-x^2\right)\,dx \;=\;2\pi\left(\tfrac{9}{2}x^2 - \tfrac{1}{4}x^4\right)\,\bigg]^3_0$

. . . . . . . . . . $=\; 2\pi\left(\frac{81}{2} - \frac{81}{4}\right) \;=\;2\pi\cdot\frac{81}{4} \;=\; \boxed{\frac{81\pi}{2}}$

6. Thats an interesting way to solve this Thanks, I will remember it

Somehow I founded math to be an interesting subject , not like it was before

7. Originally Posted by Revy
Thats an interesting way to solve this Thanks, I will remember it

Somehow I founded math to be an interesting subject , not like it was before

(this is't much connected to example in this tread but it's more for general integration )

lol you can alway check your work, just solve it without Cylindrical coordinates (to much work there), but you can do let's say with triple integral, or changing order of integration when calculating some surface, you always will get same result but as i said you get good practice and check you results (and figure out which type is more convenient to use in which case, so you will immediately know what to do with which type of integration..)

good luck

8. change the order of integration

Question:
According to
$\displaystyle \int _{b}^{a} dx \int _d ^{c} f(x,y) \;dy = \displaystyle \int _{d}^{c} dy \int _b ^{a} f(x,y) \;dx$,

if I solve
$\displaystyle \int _{0}^{2\pi } d\theta \int _0 ^{3} (9- r^2 ) r \;dr$
as
$\displaystyle \int _{0}^{3} dr \int _0 ^{2\pi } (9- r^2 ) r \;d\theta$
I should get same answer. Right?

Ah... sorry, this was answered at preveus post

9. Originally Posted by Revy
Question:
if I solve
$\displaystyle \int _{0}^{2\pi } d\theta \int _0 ^{3} (9- r^2 ) r \;dr$
as
$\displaystyle \int _{0}^{3} dr \int _0 ^{2\pi } (9- r^2 ) r \;d\theta$
I should get same answer. Right?
i this case you can do it like that, but in general can't (can but different limits ) ... look at this for example :

$\displaystyle \int_1 ^e dx \int _0 ^{\ln{x}} f(x,y) dy$

if you are to change the order of integration it's going to be

$\displaystyle \int_0 ^1 dy \int _{e^y} ^{e} f(x,y) dx$

so as you see there are going to be different limits

look at the graph, and you will conclude that integrating first on x or y have difference on how do you put your limits

10. How come you changed
$\displaystyle \int_1 ^e dx \int _0 ^{\ln{x}} f(x,y) dy$
order to
$\displaystyle \int_0 ^1 dy \int _{e^y} ^{e} f(x,y) dx$
and not to
$\displaystyle \int_0 ^{\ln{x}} dy \int _{1} ^{e} f(x,y) dx$
?

11. Originally Posted by Revy
How come you changed
$\displaystyle \int_1 ^e dx \int _0 ^{\ln{x}} f(x,y) dy$
order to
$\displaystyle \int_0 ^1 dx \int _{e^y} ^{e} f(x,y) dy$
and not to
$\displaystyle \int_0 ^{\ln{x}} dx \int _{1} ^{e} f(x,y) dy$
?
look at the graph ...

first time we go from 0 to ln(x) on y axis ... and from 1 to e on x axis

meaning

$1\leq x\leq e$

$0\leq y\leq \ln{x}$

to change order of integration just look at that same image but rotated to the left by 90° and try than

12. here you go even more simple example (can't get along with use of English) so best thing is to show how it's done

$\displaystyle \int_0 ^1 dx \int _0 ^x f(x,y) dy$

let's look at this example (check image attached)

we are integrating (look at y axis) from zero to that line y=x so it means that limits for y are zero and x okay ?
and now look how are we integrating on x axis. from zero to one so limits are zero and one

to change order of integration (draw everything until you get familiar with calculating surface or volumes using integrals) and now rotate that image to the left by 90 ° (until you get it this is easiest way)

now we are going to look how are we integrating on x axis down limit is line y=x but because we are integrating on x we are going to express that equation of the line y=x as x=y so it means that x will change from y to the one... now look at the how integration goes on y axis (intersection of the line and that vertical asymptote x=1 is in the point y=1 ) so it means that y will change from 0 to 1 and finally you get :

$\displaystyle \int _0 ^1 dy \int _y ^1 f(x,y)dx$

13. If it's correct, then I somehow got it from previous example
$\displaystyle\int_0^1dy\int_{\sqrt{y}}^1f(x,y)dx=\ int_{\sqrt{-x}}^{-1}dx\int_0^1f(x,y)dy$
we take negative $x$ from graph, so $-x$ make it positive and lets to extract root to find $y$ in graph. Or smth like that

14. Originally Posted by Revy
If it's correct, then I somehow got it from previous example
$\displaystyle\int_0^1dy\int_{\sqrt{y}}^1f(x,y)dx=\ int_{\sqrt{-x}}^{-1}dx\int_0^1f(x,y)dy$
we take negative $x$ from graph, so $-x$ make it positive and lets to extract root to find $y$ in graph. Or smth like that
no

if you pay more attention you'll see that if done like that you will after "solving" integral end up with some value ad x-es

go back to the post #9 and

$\displaystyle \int_1 ^e dx \int _0 ^{\ln{x}} f(x,y) dy = \displaystyle \int_0 ^1 dy \int _{e^y} ^{e} f(x,y) dx$

point is that here first we have had numbers as limits for x axis , and function as limit for the y axis Now if you rotate that drawing you will look at it as nothing have changed so u look how does you y changes it goes from 0 to 1 because at value of one function ln(x) is intersecting with that x=e asymptote and now x will have function for limits e^y and number e .... why is that ? well first time you have one integral with number limits and second one with expressed based on the that one that is limited with numbers ...

first time x goes from 0 to e . now if you look at the drawing you will see that y goes from 0 to ln(x) == function of x

that function is $y= \ln {x}$ so now when changing order of integration we would need to express that function based as x = ... so that can be put in limits of integral for x

$y= \ln{x}$ multiply with e and you will have $x= e^y$

does that make to you any sense ? as for that second one (with that line first time you look at it as y=x but when changing order of integration you look at that line as equation x=y )

now as far as that integral that you wrote goes ...

$\displaystyle\int_0^1dy\int_{\sqrt{y}}^1f(x,y)dx$

check image attached

so you have that $\sqrt{y}$ which is actually $x^2 = y$ so when changed order of integration it will be

$\displaystyle\int_0^1dx\int_{0}^{x^2}f(x,y)dx$