Question

Alrighty then, another question:

Calculate the volume generated when the region bounded by the curve y = 25 - x^2, the lines x = 3, y = 0 is rotated about the y-axis.

(sorry I can't provide a diagram D:)

Answer 1

Here is the plot of y=25-x^2
http://www.wolframalpha.com/input/?i=Plot%5B25-x^2%5D

Answer 2

are you sure your equations are right because it won't be a close region

Answer 3

it is a closed region. the equations given are y = 25 - x^2, x = 3, y = 0.

Answer 4

oh, yeah I didn't consider x=3

Answer 5

So it is much easier to integrate w.r.t to 'x' then with 'y'; so we have to use cylindrical method.
2pi r h delta x
R= x
h= 25-x^2
\[2\pi\int\limits\limits_{0}^{3}25-x^2(x)dx\]
\[2\pi\int\limits\limits_{0}^{3}25x-x^3(x)dx\]

Answer 6

final answer being?

Answer 7

dude wat about the left hand side of y axis??

Answer 8

is the answer
5000*pi/9 ??

Answer 9

no. lol.

Answer 10

thats a shame!!

Answer 11

is it 264*pi ??

Answer 12

\[2\pi(\int\limits_{0}^{5}25x-x^3 -\int\limits_{3}^{5}25x-x^3)\]

Answer 13

nope, siriam. sorry :/

Final answer imran?

Answer 14

(369 pi)/2

Answer 15

?

Answer 16

nope :/ Should I just post the answer up ? XD

Answer 17

625pi/2 ?

Answer 18

no :/

Aight, it's \[128\pi\]

Answer 19

how?

Answer 20

alrighty:

Consider \[y = 25 - x^2\]
\[x^2 = 25 - y\]
\[x = \sqrt{25-y}\]

At x = 3:
y = 25 - 9
y = 16.

Therefore:

Volume = \[\int\limits_{0}^{16} \pi (16-y) dy\] (integrating wrt y)
=\[\pi [16(16) - (16)^2 \div 2] - [16(0) - 0]\]
= \[128\pi\]

Answer 21

Alrighty then, I guess this place isn't right for me. Thanks anyways ^_^

Answer 22

how did u get that equation for volume?

Answer 23

He used washer method

Answer 24

sorry bare wid me coz i just read washers methd on net
but dont we get 16-y when rotated about x axis

Answer 25

it's rotated about y-axis.

Answer 26

yeah so dont we get something minus x coz in washers method the volume was
given by pi*(R^2-r^2)w
where R & r were distances of edges of rectangle rotated perpendicular to axis of rotation
here 16-y looks like some quantity parallel to it

Answer 27

The area of the cross section , in general form, is : \[\int\limits_{a}^{b} (y- y1) dx\]

Thus, the Volume (since it will be a cylinder) will be (in this case):

\[\int\limits\limits_{0}^{16} \pi (\sqrt{25-y}^2-3^2)dy\]

Answer 28

ok i get it
but then what about the volume of the part above y=16
the parabola touches y axis at (0,25)
it goes upto 25
the part above 16 is not a cylinder.

Answer 29

the region ENCLOSED by the equations y = 25 - x^2, x= 3, y = 0. pays to read the question =) It's only asking that specific region.

Answer 30

how do these equations eliminate the part above y=16 ??!

Answer 31

graph them out. :)

Answer 32

i did
still the same question

Answer 33

cmon how do u not see the part above y=16

Answer 34

the parabola is concave downwards, which restricts the region to only below the parabola. x =3 restricts one side, the parabola restricts the other. y = 0 restricts to the x-axis, giving you only up to the point where the parabola and x=3 intersect.

Answer 35

ahhh, no, wrong region, lol. You're making the region bound by x = 0, see? Going by your graph, the restrictions would be the parabola, x =3, y = 0, and x = 0.

Answer 36

is this right

Answer 37

no no, the first diagram was correct, except the region shifted to the right, underneath the parabola. Though I guess technically that would be correct, I find nothing wrong with it, the question specified the rightmost region. Sorry I couldn't provide the diagram lol :/

Answer 38

k fine
this is probably the best question i attempted here
thanks a lot for your patience