Question

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??

Answer 1

is it really a riemann sum?
if so, what is n?
or is this just finding the volume by integration?

Answer 2

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

Answer 3

yikes, okay then...

Answer 4

well which one d you want to use, midpoint, or one of the sides?

Answer 5

id rather use midpoint

Answer 6

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 \(x_i^*\)...

Answer 7

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

Answer 8

so the whole volume is given by the sum\[V\approx \sum_{i=1}^{n}A(x_i^*)\Delta x=\frac12\sum_{i=1}^{16}(x_i^{1/2}-x)^2\]

Answer 9

if you are using the midpoint rule you have\[x_1=\frac12,x_2=\frac32,...,x_{16}=\frac{31}2\]or just\[x_i^*=i-\frac12\]

Answer 10

what is the i? i somewhat understand except the i

Answer 11

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

Answer 12

\[x_n\]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

Answer 13

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

Answer 14

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?

Answer 15

yes i noticed that, y=-x

Answer 16

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

Answer 17

yes i got the sqrt(x)+x and the midpoint somewhat

Answer 18

just so we have an image to reference...