Thread: Need help w/ Riemann Sums+Antiderivatives please

1. Need help w/ Riemann Sums+Antiderivatives please

Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions

2. Originally Posted by MrJigsaw
Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions

Your question makes no sense: you give a one-variable function and you ask for its volume...what volume?!? You could ask about the area enclosed between the function's graph and the x-axis, or even about the volume of the 3-dimensional body created by revolving that graph around the x-axis or any other line, but just as it is the question makes no sense.

Tonio

3. Originally Posted by MrJigsaw
Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions
I'm going to assume by volume you mean $\displaystyle 2$ dimensional volume, aka area.

Are you familiar with integration by parts?

If so take $\displaystyle u=\log(x),\;\; dv=dx \implies du=\frac{dx}{x},\;\;v=x$

Thus $\displaystyle \int \log(x)dx = x\log(x)-\int x\frac{dx}{x} = x\log(x)-x+C$

4. My apologies. I meant for that function to be rotated around the x-axis. Hope that clarifies.

No, I can't say I am familiar with integration by parts. My teacher said that I may need to find the opposite by switching x and y and finding a new equation involving ln but she didn't fully explain.

5. Originally Posted by MrJigsaw
Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions
Are you talking about the area? Use the change of base formula and you get

$\displaystyle \int (\ln{x} - \ln{10} + 2)dx$

Use integration by parts for the first term

6. Originally Posted by MrJigsaw
My apologies. I meant for that function to be rotated around the x-axis. Hope that clarifies.

No, I can't say I am familiar with integration by parts. My teacher said that I may need to find the opposite by switching x and y and finding a new equation involving ln but she didn't fully explain.
Oh, in that case we get $\displaystyle V = \pi\int_{.2}^7 (\log(x)-2)^2dx = \pi\int_{.2}^7 (\log^2(x)-4\log(x)+4)dx$.

Would you be able to evaluate this integral?

7. With my calculator yes lol. But I need to show work. I haven't been able to find any help with integrating log

8. Originally Posted by MrJigsaw
My apologies. I meant for that function to be rotated around the x-axis. Hope that clarifies.

No, I can't say I am familiar with integration by parts. My teacher said that I may need to find the opposite by switching x and y and finding a new equation involving ln but she didn't fully explain.
Let's try it this way then:

$\displaystyle y=\log(x)+2\implies x=e^{y-2}$ and our new interval is $\displaystyle [\log(.2)+2,\;\log(7)+2]$

Now we want to rotate our new equation around the y-axis.

Can you manage to do this?

9. So I start with $\displaystyle V = \pi\int_{log(.2)+2}^{log(7)+2} (e^{y-2} )dy$ right?

Then that becomes $\displaystyle V = \pi\ ((y-2)(e^{y-2}))$ from log(.2)+2 to log(7)+2?

Then just plug in the top and bottom and subtract?

10. Originally Posted by MrJigsaw
So I start with $\displaystyle V = \pi\int_{log(.2)+2}^{log(7)+2} (e^{y-2} )dy$ right?

Then that becomes $\displaystyle V = \pi\ ((y-2)(e^{y-2}))$ from log(.2)+2 to log(7)+2?

Then just plug in the top and bottom and subtract?
No, there is a different formula for rotating about the y-axis.

11. Oh right, this would be washer method. So I just make it 2pi? Or did i forget something else? Isn't it outer radius minus inner? And the inner is 0

12. Please refer to the figure below to help picture things geometrically.

For the interval $\displaystyle x=0$ to $\displaystyle x=\log(.2)+2$ we need to calculate the volume of the cylinder with radius $\displaystyle \log(.2)+2$ and height $\displaystyle 6.8$.

For the interval $\displaystyle x=\log(.2)+2$ to $\displaystyle x=\log(7)+2$ we need to apply the shell method (I think that's what it's called).

So looking at the small sliver from $\displaystyle y=e^{x-2}$ to $\displaystyle y=7$ and thickness $\displaystyle \Delta x$, we want to revolve that around the y-axis and calculate the volume of the thin cylindrical shell which is about $\displaystyle 2\pi x h \Delta x$ (note I'm not being very rigorous here). Next see that $\displaystyle h=7-e^{x-2}$.

Let's call this part of the volume $\displaystyle V_2$:

$\displaystyle V_2\approx \sum 2\pi x (7-e^{x-2})\Delta x$

As $\displaystyle \Delta x\to 0$ our sum turns into the integral $\displaystyle V_2 = 2\pi \int_{\log(.2)+2}^{\log(7)+2} x\cdot(7-e^{x-2})dx$

Therefore $\displaystyle V = \pi\cdot6.8\cdot(\log(.2)+2)^2+2\pi \int_{\log(.2)+2}^{\log(7)+2} x\cdot(7-e^{x-2})dx \approx 211.977153$

13. Alright thanks a lot everyone. I got it now

14. Originally Posted by MrJigsaw
Alright thanks a lot everyone. I got it now
If you aren't familiar with integration by parts, how are you going to be able to evaluate the final integral?