Need help in integral area involve 3d objects

- anonymous

Need help in integral area involve 3d objects

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- schrodinger

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- anonymous

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- anonymous

what's your question?

- anonymous

ah I see those are multiple choices...

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- anonymous

did you get the 'sketch the region' one?

- anonymous

basically just asking you to pick which of those areas is bounded by y=5, x=4 and the function...

- anonymous

you there?

- anonymous

the first choice is bounded by y=10, x=4 and the function...
the second choice is bounded by y=10, x=0, x=4 and the function...

- anonymous

yes i am here..

- anonymous

I know the answer for the first picture already, not sure about the 2nd one @Algebraic!

- anonymous

For the first picture, i think it is the last one
for the 2nd picture, i am thinking either 1st or 4th one , what do u think ?

- anonymous

naw, the last picture is bounded by x=0 y=0 and the function...

- anonymous

(and x=4)

- anonymous

so for the first picture, u also think is the last one?

- anonymous

for my function, x is not 0 u mean?

- anonymous

just trace around each of those red regions and see what functions or lines your tracing...

- anonymous

what lines do you trace for the last choice?

- anonymous

|dw:1352777609189:dw|

- anonymous

yea, that's what i have too. dont really know how i got that..but yeah..
Do you possibly know how to volume the volume of this?

- anonymous

|dw:1352777630991:dw|

- anonymous

|dw:1352777716676:dw|

- anonymous

dont u need to integrate the area function to volume , and then something to do with the height..?

- anonymous

so... is that the region described in your problem statement??

- anonymous

yes

- anonymous

really?
your problem statement makes no mention of x=0 or y=0 ...

- anonymous

it does mention y=5 however...

- anonymous

well, but it says rotate about y= 10, and y= 5

- anonymous

naw..

- anonymous

then u can kinda know there is a hole in between

- anonymous

umm..wait..so which picture do you think it is for the first set of picture? i thought we have the same answer

- anonymous

the region is bounded by x=4 y=5 and y=5e^-x

- anonymous

like I said trace the boundary of the regions you have been given to find which one works with that description..

- anonymous

there's no point in guessing here... even though this is multiple guess

- anonymous

i still think is the last one..

- anonymous

that's why i want to know which one u think it is

- anonymous

why is it the last one..

- anonymous

are you talking about the first part or the second part?

- anonymous

first part ..

- anonymous

because if you do the first part correctly, the answer to the second part should be obvious

- anonymous

ok..

- anonymous

so do you agree that it is either the bottom left or bottom right picture is correct
due to the fact that given y=5

- anonymous

then why are you saying it's the last region?

- anonymous

why is it not the last picture ..lol |dw:1352778179844:dw|

- anonymous

did you see the sketches where I traced the bounding functions??

- anonymous

what are they??

- anonymous

the curve is y= 5e^-x , y =5 , x=4

- anonymous

what functions bound the region in the last picture?

- anonymous

hint: I already did it for you.

- anonymous

x=0, y=0..which i dont really get u ..o_o

- anonymous

trace the boundary

- anonymous

|dw:1352778560923:dw|

- anonymous

what's that portion?

- anonymous

say "y=0"

- anonymous

|dw:1352778627593:dw|

- anonymous

what's that portion?

- anonymous

are you trying to say that radius doesnt bound to x=0?

- anonymous

which it does indeed make sense when i look in the 2nd part

- anonymous

I'd focus on the first part, if I were you..

- anonymous

picking which region is bounded by y=5, x=4 and y= 5e^-x

- anonymous

i'm not sure what u mean by bounded..do you mean by starting from thereï¼Ÿ

- anonymous

trace the boundary of each region

- anonymous

like I did... for the last region pictured in your choices...

- anonymous

if i do what u did, then the answer will be the bottom left

- anonymous

yes

- anonymous

but my question with that picture is.. don't u need to do 5- 5e^-t in order to get tht area?

- anonymous

not necessarily..

- anonymous

you'll be using washers probably, since you haven't learned about shells yet (I assume)...

- anonymous

so the inner radius is 5 and the outer radius is 10-5e^-x

- anonymous

yea i didnt yet.. i didnt even learn these stuff much when my assignment is due 2 days after..

- anonymous

ya, i do get that part

- anonymous

|dw:1352779152955:dw|

- anonymous

is that region not what the last picture look like?

- anonymous

for the second part... yep.

- anonymous

can i ask what does "region " mean..

- anonymous

the red areas in the first part.

- anonymous

yea..what does that red area mean

- anonymous

is it like the path of rotation or something?

- anonymous

no.

- anonymous

it's the thing you're sweeping around the specified axis in order to generate a solid

- anonymous

|dw:1352779636199:dw|

- anonymous

take that region as an example... it's a rectangle...

- anonymous

rotate about the x axis and you sweep out a cylinder..

- anonymous

so it's kinda you are sweeping that region, and that sweeping motion generate the solid figure?

- anonymous

|dw:1352779702320:dw|

- anonymous

yes, exactly.

- anonymous

okay thank you so so much!!!!!!

- anonymous

Do you still have time to guide me on filding the volume?

- anonymous

the second multiple choice question? or a new question?

- anonymous

it's the same question. but it asks to find the volume

- anonymous

cuz finding the volume is actually the first part..but the graphing part seems easier so i skipped to that first ..

- anonymous

sure... we are using the washers in the sketch (and in the last choice to the second question)

- anonymous

inner radius is 5, outer radius is 10 - 5e^-x

- anonymous

area of a washer will be pi( (10 - 5e^-x)^2 - 5^2 )

- anonymous

yea..that's all i have too.. not sure how to integrate that..

- anonymous

\[\pi \int\limits_{ }^{} (10-5e^{-x})^2 - 5^2 dx\]

- anonymous

\[\pi \int\limits_{0}^{4} 75 -100e^{-x} +25e^{-2x} dx\]

- anonymous

that should be pretty easy to integrate...

- anonymous

oh ok..i didnt combine them.. ok thanks, i will try that out :)

- anonymous

@Algebraic! umm.. how can i leave it as exact numbers..?

- anonymous

i got pi ((300+10 e^-4 - 25/2 e^-8)- (10-25/2))

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