anonymous
  • anonymous
Need help with riemann sum, i have the y=sqrt(x) y=-x, x=16. the cross sections are right isosceles triangles perpendicular to the x-axis. help with volume??
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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TuringTest
  • TuringTest
is it really a riemann sum? if so, what is n? or is this just finding the volume by integration?
anonymous
  • anonymous
i was planning to cut it into 16 pieces, but the assignment says to show volume of each layer and indicate whether im using left or right or midpoint riemann sum. i got 1156.267 for the volume
TuringTest
  • TuringTest
yikes, okay then...

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TuringTest
  • TuringTest
well which one d you want to use, midpoint, or one of the sides?
anonymous
  • anonymous
id rather use midpoint
TuringTest
  • TuringTest
ok, so how about redefining a function for the length of the base in terms of \(x_i^*\) that will be \(b(x_i^*)=x_i^{1/2}-x_i\) Now we need a function for the area of a given \(x_i^*\)...
TuringTest
  • TuringTest
triangle area=bh/2 since it is isosceles b=h we have\[A(x_i^*)=\frac12(b(x))^2=\frac12(x_i^{1/2}-x_i)^2\]multiplying that by \(\Delta x=1\) (if \(n=16\)) gives the volume of the each 'slice' of triangle
TuringTest
  • TuringTest
so the whole volume is given by the sum\[V\approx \sum_{i=1}^{n}A(x_i^*)\Delta x=\frac12\sum_{i=1}^{16}(x_i^{1/2}-x)^2\]
TuringTest
  • TuringTest
if you are using the midpoint rule you have\[x_1=\frac12,x_2=\frac32,...,x_{16}=\frac{31}2\]or just\[x_i^*=i-\frac12\]
anonymous
  • anonymous
what is the i? i somewhat understand except the i
TuringTest
  • TuringTest
the "index" of the terms it is just the number in the subscript that we use to designate which value of x we are talking about
TuringTest
  • TuringTest
\[x_n\]is the \(n\)th term in the sequence ...don't let all the different names fool you, it is just a "counter" if the last equation I wrote\[x_i^*=i-\frac12\]confuses you just ask yourself if it works if you are using the midpoint rule is the first term (the i=1 terms) not 1/2 ? is the second not 3/2 you can see that the \(i\)th value of x in this particular case can be described by the formula I wrote above
anonymous
  • anonymous
could you calculate the area and volume of one slice for me? for an example, i have to find the area and volume for each individual slice
TuringTest
  • TuringTest
I made a typo above and had the base \(b(x) =\sqrt x-x\) when it should have been \(b(x)=\sqrt x+x\), each face should have an area of \[A(x)=\frac12bh=\frac12(\sqrt x+x)^2\]do you see that?
anonymous
  • anonymous
yes i noticed that, y=-x
TuringTest
  • TuringTest
my bad :/ but you do you see my reasoning so far about the area of each face of a given triangle? please be honest, don't just say yes :P
anonymous
  • anonymous
yes i got the sqrt(x)+x and the midpoint somewhat
TuringTest
  • TuringTest
|dw:1332714608011:dw|just so we have an image to reference...
TuringTest
  • TuringTest
the base of each triangle is this difference between the two functions|dw:1332714802711:dw|and each triangle is right-iscoceles so it looks like this|dw:1332714885084:dw|base=height because it is isosceles, so we have the area as...
TuringTest
  • TuringTest
\[A(x)=\frac12bh=\frac12[b(x)]^2=\frac12(\sqrt x+x)^2\]this is the area of the face of a given triangle at a point x so far how are we?
anonymous
  • anonymous
i understand
TuringTest
  • TuringTest
good, so then all you need is to multiply by the thickness of each slice, which is \(\Delta x\) and you get the volume of that 'slice' of triangle you said you wanted to use 16 pieces, so what is your \(\Delta x\) ?
TuringTest
  • TuringTest
...because\[V(x)=A(x)\Delta x\]in case I failed to make that clear ...so again, what is your \(\Delta x\) ?
anonymous
  • anonymous
\[Deltax=1\]
anonymous
  • anonymous
i think
TuringTest
  • TuringTest
yep :D so you are done, aside from choosing a point to evaluate
anonymous
  • anonymous
i still dont under stand how i would get area and volume from jsut one slice
TuringTest
  • TuringTest
just choose one particular x in the formula I derived for \(A(x)\) the volume of a slice is\[V(x_i)=A(x_i)\Delta x=\frac12(\sqrt x_i+x_i)^2(1)=\frac12(\sqrt x_i+x_i)^2\]where \(x_i\) is some point random point in out interval since we are using the midpoint rule, I demonstrated that\[x_i=i-\frac12\]so pick any \(i\) between 1 and 16, use that to get a particular x, then plug it into the formula for volume.
anonymous
  • anonymous
so i=1 would be \[1/2(\sqrt{1-.5)}+1-.5)^2\]
TuringTest
  • TuringTest
...for instance, using the midpoint rule, the first \(x_i^*\) is\[x_i=i-\frac12=\frac12\]so the volume there is\[V(x_1)=A(x_1)\Delta x=(\sqrt{\frac12}+\frac12)^2\]
TuringTest
  • TuringTest
which is exactly what you have, yes :)
TuringTest
  • TuringTest
that shoulde be \[x_1=1-\frac12=12\]at the start of my post; typo*
TuringTest
  • TuringTest
gr...\[x_1=1-\frac12=\frac12\]
anonymous
  • anonymous
and that is volume so would area be that same answer for volume?
TuringTest
  • TuringTest
yeah, because you chose to break it into n=16 pieces, so in \(this\) particular case \[\Delta x=1\implies V(x)=A(x)\Delta x=A(x)(1)=A(x)\]now if you had to break this into 32 pieces, \(\Delta x\) would be 1/2, so we would need to multiply every are by 1/2 to get the volume just remember: volume=area of base X thickness
TuringTest
  • TuringTest
every area
TuringTest
  • TuringTest
*
anonymous
  • anonymous
did you make a mistake you put (sqwrt(1/2)-1/2)^2, shouldnt there be a 1/2 in front, be cause of the area of a triangle??
TuringTest
  • TuringTest
for x_1 ? yes I did :)\[V(x_1)=A(x_1)\Delta x=\frac12(\sqrt{\frac12}+\frac12)^2\]
TuringTest
  • TuringTest
but you made a mistake and put a minus, so there :P lol
anonymous
  • anonymous
haha thanks for your help, i understand
TuringTest
  • TuringTest
awesome glad to help :D

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