Question

Find the volume of the solid obtained by rotating the region bounded by the given curve about the y-axis.

y=3xe^-x y=0 x=3 about the y=axis

I was able to get the graph and all I really need is someone to explain how to set up the integral. Thanks for your help!

Answer 1

Do yo mean,
\[y=3xe^{-x}\] or
\[y=3e^{-x}\]?

Answer 2

the first one

Answer 3

with one dimensional integration?

Answer 4

yes

Answer 5

Sorry, I don't see an easy way to solve this in one dimension.

Answer 6

Around the y axis. Around the x axis it should be easy.

Answer 7

the question says around the y-axis is it easier as a 3D figure?

Answer 8

As a 3D figure, integrating in three dimensions, I think it should be easier. The problem I see when you integrate around y axis is that you need the funtion f(y), but the equation you give don't let us put x in function of y, so we cannot find an easy expression for f(y). Is this problem from a book or could you tell me from what course is?

Answer 9

its from calc 2

Answer 10

What textbook do you use?

Answer 11

Calculus Early Transcendentals 7E by james stewart

Answer 12

Ok, I will check it. May be we'll have some luck.

Answer 13

Could you please indicate me the chapter?

Answer 14

6.3

Answer 15

Ok, I completely forget about it. But it is easy. It is integration using cylindrical shells. The volume generated by a function f(x) (like the one they give you, where you cannot put x as a function of y), is,
\[V=\int_a^b2\pi x f(x)dx\]So in this case,
\[V=\int_0^32\pi x (3xe^{-x})dx=6\pi\int_0^3 x^2e^{-x}dx\]
Could you integrate it?

Answer 16

i think so thanks! I was just really unsure of the e

Answer 17

Ok, only to check it I obtain the value, and find,
\[V=6 \left(2-\frac{17}{e^3}\right) \pi\]
I hope it helps you.

Answer 18

it did! Could you help me with one other problem?

Answer 19

I'll try it :)

Answer 20

The integral represents the volume of a solid. Describe the solid.
\[2\pi \int\limits_{0}^{6} \frac{ 7y }{ 7+y^2 } dy\]
The solid is obtained by rotating the region \[o \le x \le ? \] \[0 \le y \le 6 \] about the x-axis using cylindrical shells.

Answer 21

I was able to figure out the y and the x-axis part but i don't see how to sole for the x part

Answer 22

To solve for the x part we need the function f(x). If you compare the integral you give with the one we use to integrate volumes around the y axis,
\[V=\pi\int f(y)^2dy\]We know that
\[x=f(y)\]In our case,
\[f(y)^2=2\frac{7y}{7+y^2}=\frac{14y}{7+y^2}\Rightarrow f(y)=\sqrt{\frac{14y}{7+y^2}}\]So,
\[x=\sqrt{\frac{14y}{7+y^2}}\]Now you should only substitute the y with value 6, and you'll obtain the required result.

Answer 23

My result is, to check,
\[2 \sqrt{\frac{21}{43}}\]

Answer 24

It says its wrong

Answer 25

\[0 \le x \le 2\sqrt{\frac{ 21 }{ 43 }}\] is not right

Answer 26

wait shouldnt V= 2pi not just pi

Answer 27

May be they don't simplify,
\[\sqrt{84/43}\]

Answer 28

Does your integral has a 2 or not? :)

Answer 29

\[V=2\pi \int\limits_{a}^{b}y f(y) dy \]

Answer 30

thats the equation the question gives

Answer 31

Ah, and the f(y) you posted before?

Answer 32

Anyways, if the problem gives you the last expression, could you write f(y)?

Answer 33

yea the integral from 0 to 6

Answer 34

i thought it was \[\sqrt{\frac{ 7 }{ x-7 }}\] but it said it was wrong

Answer 35

Ok, I think it should be this way,
\[V=2\pi\int_0^6yf(y)dy\Rightarrow x=\frac{7}{7+y^2}\Rightarrow x=\frac{7}{7+36}=7/43\]

Answer 36

Ok, I think it should be this way,
\[V=2\pi\int_0^6yf(y)dy\Rightarrow x=\frac{7}{7+y^2}\Rightarrow x=\frac{7}{7+36}=7/43\]

Answer 37

got it wrong the answer is actually 7/7+y^2 not suppose to plug in the 6 but thanks for your help!!

Answer 38

Oh, of course. I was thinking in a numeric limit! But they ask you for the function! Sorry. ;)

Answer 39

its fine thanks for your help :) I only have one more problem to figure out so if i get this one i think ill be ok lol

Answer 40

Well, post it and we'll try :)

Answer 41

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method. \[y=-x^2+8x-15 \] y=0 about the x-axis

Answer 42

What is the other curve and where are the limits of integration?

Answer 43

thats all the question gives :/

Answer 44

http://www.wolframalpha.com/input/?i=graph+y%3D-x%5E2+%2B8x+-15

Answer 45

since y=0 the intersection points are (0,3) and (0,5)

Answer 46

True. I just read curves and was thinking in another parabola XD.

Answer 47

Then it should be easy,
\[V=\pi\int_3^5 (-x^2+8x-15)^2dx\]

Answer 48

Something like this,
http://www.wolframalpha.com/input/?i=Integrate%5BPi+%28-x%5E2+%2B8x+-15%29%5E2%2C%7Bx%2C3%2C5%7D%5D

Answer 49

that was so simple i guess i was over thinking it haha

Answer 50

Sometimes happens ;)