# [SOLVED]Riemann sum to estimate volume

#### Houdini

##### New member
Question:
A solid has a rectangular base that lies in the first quadrant and is bounded by the x and y-axes and the lines x=2, y=1. The height of the solid above point (x,y) is 1+3x. Find the Riemann approximation of the solid.

Solution:

I already know that the solution is $$\displaystyle \sum_{i=1}^{n} \frac{2}{n} \left(1+\frac{6i}{n} \right)$$. What I don't see is why it's 1+(6i)/n and not 1+(3i)/n. Volume can be generalized to be the area of the base times the height, so for this problem I have something like $$\displaystyle x*y*f(x)$$. Of course x is changing so I must rewrite this.

For any partition where we approximate the volume between x_1 and x_2 the length will be $$\displaystyle \Delta x=\frac{2}{n}$$ The y value is a constant 1, so won't need to be written explicitly as far as I can see. I know this part is incorrect but it seems to me that the height should be $$\displaystyle 1+\frac{3i}{n}$$, but I know that since we haven't defined where $$\displaystyle x_i$$ is in each partition (it could be the left value, middle value, right value or anywhere) then I'm really stuck here.

EDIT: Now that I think about it more since we haven't defined what n is we don't know what i/n either and i/n will change according to how many partitions we take. So am I correct in thinking that $$\displaystyle f(x_i)=f(i\Delta x)$$? If so this makes sense now.

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#### CaptainBlack

##### Well-known member
Question:
A solid has a rectangular base that lies in the first quadrant and is bounded by the x and y-axes and the lines x=2, y=1. The height of the solid above point (x,y) is 1+3x. Find the Riemann approximation of the solid.

Solution:

I already know that the solution is $$\displaystyle \sum_{i=1}^{n} \frac{2}{n} \left(1+\frac{6i}{n} \right)$$. What I don't see is why it's 1+(6i)/n and not 1+(3i)/n. Volume can be generalized to be the area of the base times the height, so for this problem I have something like $$\displaystyle x*y*f(x)$$. Of course x is changing so I must rewrite this.

For any partition where we approximate the volume between x_1 and x_2 the length will be $$\displaystyle \Delta x=\frac{2}{n}$$ The height is a constant 1, so won't need to be written explicitly as far as I can see. I know this part is incorrect but it seems to me that the height should be $$\displaystyle 1+\frac{3i}{n}$$, but I know that since we haven't defined where $$\displaystyle x_i$$ is in each partition (it could be the left value, middle value, right value or anywhere) then I'm really stuck here.

EDIT: Now that I think about it more since we haven't defined what n is we don't know what i/n either and i/n will change according to how many partitions we take. So am I correct in thinking that $$\displaystyle f(x_i)=f(i\Delta x)$$? If so this makes sense now.
This is just a one dimensional integral/Riemann sum problem in disguise, since the height is a function of $$x$$ only and the interval for $$y$$ is of width $$1$$, so we seek:
$\int_{x=0}^2 (1+3x) \; dx$
We divide the $$x$$-interval $$[0,2]$$ into $$n$$ strips each of width $$2/n$$, then the area of the $$i$$-th strip is approximately:
$A_i=\frac{2}{n} \times \left(1+3 \times \frac{2\times i}{n}\right)$
where we are using the $$x$$ value of the right edge of the $$i$$-th strip ($$(2 \times i)/n$$ ) as our approximate value of $$x$$ in the formula for the height of the for the strip.

Then the right Riemann sum is:
$S_n=\sum_{i=1}^{n}A_i={\Large{\sum_{i=0}^{n-1}}}\left[ \frac{2}{n} \times \left(1+3 \times \frac{2\times i}{n}\right)\right]$.

CB

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#### Houdini

##### New member
Almost makes perfect sense, just would like clarification about the $$\displaystyle f(x_i)=f(i \times \Delta x)$$ part. If that's true than I see $$\displaystyle f(x_i)$$ as $$\displaystyle \left( 1+ 3 \left[ i \times \frac{2}{n} \right] \right)$$. I know that's just rewriting what you posted but writing it this way seems to confirm my stated assumption.

That's well put that this is a disguised single variable integral too.

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