anonymous
  • anonymous
Need help in integral area involve 3d objects
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
what's your question?
anonymous
  • anonymous
ah I see those are multiple choices...

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anonymous
  • anonymous
did you get the 'sketch the region' one?
anonymous
  • anonymous
basically just asking you to pick which of those areas is bounded by y=5, x=4 and the function...
anonymous
  • anonymous
you there?
anonymous
  • 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
  • anonymous
yes i am here..
anonymous
  • anonymous
I know the answer for the first picture already, not sure about the 2nd one @Algebraic!
anonymous
  • 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
  • anonymous
naw, the last picture is bounded by x=0 y=0 and the function...
anonymous
  • anonymous
(and x=4)
anonymous
  • anonymous
so for the first picture, u also think is the last one?
anonymous
  • anonymous
for my function, x is not 0 u mean?
anonymous
  • anonymous
just trace around each of those red regions and see what functions or lines your tracing...
anonymous
  • anonymous
what lines do you trace for the last choice?
anonymous
  • anonymous
|dw:1352777609189:dw|
anonymous
  • 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
  • anonymous
|dw:1352777630991:dw|
anonymous
  • anonymous
|dw:1352777716676:dw|
anonymous
  • anonymous
dont u need to integrate the area function to volume , and then something to do with the height..?
anonymous
  • anonymous
so... is that the region described in your problem statement??
anonymous
  • anonymous
yes
anonymous
  • anonymous
really? your problem statement makes no mention of x=0 or y=0 ...
anonymous
  • anonymous
it does mention y=5 however...
anonymous
  • anonymous
well, but it says rotate about y= 10, and y= 5
anonymous
  • anonymous
naw..
anonymous
  • anonymous
then u can kinda know there is a hole in between
anonymous
  • 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
  • anonymous
the region is bounded by x=4 y=5 and y=5e^-x
anonymous
  • anonymous
like I said trace the boundary of the regions you have been given to find which one works with that description..
anonymous
  • anonymous
there's no point in guessing here... even though this is multiple guess
anonymous
  • anonymous
i still think is the last one..
anonymous
  • anonymous
that's why i want to know which one u think it is
anonymous
  • anonymous
why is it the last one..
anonymous
  • anonymous
are you talking about the first part or the second part?
anonymous
  • anonymous
first part ..
anonymous
  • anonymous
because if you do the first part correctly, the answer to the second part should be obvious
anonymous
  • anonymous
ok..
anonymous
  • 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
  • anonymous
then why are you saying it's the last region?
anonymous
  • anonymous
why is it not the last picture ..lol |dw:1352778179844:dw|
anonymous
  • anonymous
did you see the sketches where I traced the bounding functions??
anonymous
  • anonymous
what are they??
anonymous
  • anonymous
the curve is y= 5e^-x , y =5 , x=4
anonymous
  • anonymous
what functions bound the region in the last picture?
anonymous
  • anonymous
hint: I already did it for you.
anonymous
  • anonymous
x=0, y=0..which i dont really get u ..o_o
anonymous
  • anonymous
trace the boundary
anonymous
  • anonymous
|dw:1352778560923:dw|
anonymous
  • anonymous
what's that portion?
anonymous
  • anonymous
say "y=0"
anonymous
  • anonymous
|dw:1352778627593:dw|
anonymous
  • anonymous
what's that portion?
anonymous
  • anonymous
are you trying to say that radius doesnt bound to x=0?
anonymous
  • anonymous
which it does indeed make sense when i look in the 2nd part
anonymous
  • anonymous
I'd focus on the first part, if I were you..
anonymous
  • anonymous
picking which region is bounded by y=5, x=4 and y= 5e^-x
anonymous
  • anonymous
i'm not sure what u mean by bounded..do you mean by starting from thereļ¼Ÿ
anonymous
  • anonymous
trace the boundary of each region
anonymous
  • anonymous
like I did... for the last region pictured in your choices...
anonymous
  • anonymous
if i do what u did, then the answer will be the bottom left
anonymous
  • anonymous
yes
anonymous
  • anonymous
but my question with that picture is.. don't u need to do 5- 5e^-t in order to get tht area?
anonymous
  • anonymous
not necessarily..
anonymous
  • anonymous
you'll be using washers probably, since you haven't learned about shells yet (I assume)...
anonymous
  • anonymous
so the inner radius is 5 and the outer radius is 10-5e^-x
anonymous
  • anonymous
yea i didnt yet.. i didnt even learn these stuff much when my assignment is due 2 days after..
anonymous
  • anonymous
ya, i do get that part
anonymous
  • anonymous
|dw:1352779152955:dw|
anonymous
  • anonymous
is that region not what the last picture look like?
anonymous
  • anonymous
for the second part... yep.
anonymous
  • anonymous
can i ask what does "region " mean..
anonymous
  • anonymous
the red areas in the first part.
anonymous
  • anonymous
yea..what does that red area mean
anonymous
  • anonymous
is it like the path of rotation or something?
anonymous
  • anonymous
no.
anonymous
  • anonymous
it's the thing you're sweeping around the specified axis in order to generate a solid
anonymous
  • anonymous
|dw:1352779636199:dw|
anonymous
  • anonymous
take that region as an example... it's a rectangle...
anonymous
  • anonymous
rotate about the x axis and you sweep out a cylinder..
anonymous
  • anonymous
so it's kinda you are sweeping that region, and that sweeping motion generate the solid figure?
anonymous
  • anonymous
|dw:1352779702320:dw|
anonymous
  • anonymous
yes, exactly.
anonymous
  • anonymous
okay thank you so so much!!!!!!
anonymous
  • anonymous
Do you still have time to guide me on filding the volume?
anonymous
  • anonymous
the second multiple choice question? or a new question?
anonymous
  • anonymous
it's the same question. but it asks to find the volume
anonymous
  • anonymous
cuz finding the volume is actually the first part..but the graphing part seems easier so i skipped to that first ..
anonymous
  • anonymous
sure... we are using the washers in the sketch (and in the last choice to the second question)
anonymous
  • anonymous
inner radius is 5, outer radius is 10 - 5e^-x
anonymous
  • anonymous
area of a washer will be pi( (10 - 5e^-x)^2 - 5^2 )
anonymous
  • anonymous
yea..that's all i have too.. not sure how to integrate that..
anonymous
  • anonymous
\[\pi \int\limits_{ }^{} (10-5e^{-x})^2 - 5^2 dx\]
anonymous
  • anonymous
\[\pi \int\limits_{0}^{4} 75 -100e^{-x} +25e^{-2x} dx\]
anonymous
  • anonymous
that should be pretty easy to integrate...
anonymous
  • anonymous
oh ok..i didnt combine them.. ok thanks, i will try that out :)
anonymous
  • anonymous
@Algebraic! umm.. how can i leave it as exact numbers..?
anonymous
  • anonymous
i got pi ((300+10 e^-4 - 25/2 e^-8)- (10-25/2))

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