Volume by Shells

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Hi.. i was wondering if i could check to see if im doing volume by shells correctly, thanks!

#1
y=x^2, y=2x about the line y=5

2 pi S y(y^.5- y/2)dy
2 pi [y^2.5 - y^3/6] from 0 to 4
= 64pi/15 units^3

#2 My answer doesnt look right, so i think this one is wrong.

x=y^3-y^4, x=0 about the line y=-2

2pi S y(y^3-y^4)dy
2pi[y^5/5-y^6/6] from 0 to 1
=pi/15 units^3

#3 I think my set up is correct (but im not sure) but i am not sure how to integrate e^(-x^2)...

y=e^(-x^2), y=0, x=0, x=1 about the y-axis

2piS e^(-x^2) from 0 to 1....??

last one!
y=1/x^2, y=0, x=1, x=2 about x=-1

2pi S x* 1/x^2 dx
2 pi [lnx] from 1 to 2
= 2pi*ln2 units^3

THANKS!! :)
 
aswimmer113 said:
Hi.. i was wondering if i could check to see if im doing volume by shells correctly, thanks!

#1
y=x^2, y=2x about the line y=5

2 pi S y(y^.5- y/2)dy
2 pi [y^2.5 - y^3/6] from 0 to 4
= 64pi/15 units^3

You forgot, you're not revolving it around the y axis, but around y=5.

\(\displaystyle \L\\2{\pi}\int_{0}^{4}(y-5)(\frac{y}{2}-\sqrt{y})dx\)

or

\(\displaystyle \L\\{\pi}\int_{0}^{2}((5-x^{2})^{2}-(5-2x)^{2})dx\)

#2 My answer doesnt look right, so i think this one is wrong.

x=y^3-y^4, x=0 about the line y=-2

2pi S (y+2)(y^3-y^4)dy
2pi[y^5/5-y^6/6] from 0 to 1
=pi/15 units^3



#3 I think my set up is correct (but im not sure) but i am not sure how to integrate e^(-x^2)...

y=e^(-x^2), y=0, x=0, x=1 about the y-axis

2piSxe^(-x^2) from 0 to 1....??

It seems I must edit due to Soroban's reminder. I, too, overlooked the obvious. :oops:

last one!
y=1/x^2, y=0, x=1, x=2 about x=-1

2pi S (x+1)* 1/x^2 dx
2 pi [lnx] from 1 to 2
= 2pi*ln2 units^3


THANKS!! :)
Click to expand...
 
Hello, aswimmer113!

You seem to forget parts of the formula:

\(\displaystyle \L\;\;V\;=\;2\pi\int^{\;\;\;b}_a\text{ (radius)(height) }dx\;\; \text{(or }dy)\)

\(\displaystyle 3)\;\;y\,=\,e^{-x^2},\;y\,=\,0,\;x\,=\,0,\,x\,=\,1\) about the y-axis
Click to expand...

\(\displaystyle \text{Radius }=\;x,\;\;\text{Height }=\;e^{-x^2}\)

\(\displaystyle \L V\;=\;2\pi\int^{\;\;\;1}_0x\cdot e^{-x^2}\,dx\)

Let \(\displaystyle u \,= \,-x^2\;\;\Rightarrow\;\;du \,= \,-2x\,dx\;\;\Rightarrow\;\;dx\,=\,-\frac{du}{2x}\)

Substitute: \(\displaystyle \L\,V \;= \;2\pi\int x\cdot e^u\cdot\left(-\frac{du}{2x}\right) \;= \;-\pi\int e^u\,du \;= \;-\pi e^u\)

Back-substitute: \(\displaystyle \L\,V\;=\;-\pi \left[ e^{-x^2}\,\right]^1_0 \;=\;-\pi\left(e^{-1}\,-\,e^0\right) \;=\;-\pi\left(\frac{1}{e}\,-\,1\right)\)

Therefore: \(\displaystyle \L\,V\;=\;-\pi\left(\frac{1\,-\,e}{e}\right)\;=\;\pi\left(\frac{e\,-\,1}{e}\right)\)

 
thanks for the help! im still confused on #1 though...

could i do

2 pi integral (y^.5-5)(y/2-5) dy ??
 
oh wait.. that doesnt make sense... i think im confusing shells with cross sections... UGH
 
okay.. im confused

so i do 2 pi S r*h

so i get (x^2-2x)*(5-x) ??
 
ah im sorry im so confused... but i have to make them Y

So... could it maybee be..

2 pi S (from 0 to 4) (5-y)(y^.5-y/2)???? Please let this be it...
 
Check out my first post. What do you see?.

Anything familiar?.
 
well, the way we have to do it it has to be dy. And you had y-5 and had an x in the y/x - x^.5 ...
 
That's just a typo. Thanks for pointing it out.

Either way I have posted will work. They're the same. Just two different methods.
 
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