Volume Integral

[Paul Heisenberg]

I ran into a problem solving this exercise. While I was able to solve the Vx correctly I failed in doing so for Vy. I tried to solve it using the yellow equation, but I doesn't result in the identical Integral as in the solution. My question is now, is the yellow equation wrong, or shouldn't I be using this on this problem or what am I doing wrong? In the solution they do a trick with Vy(a)=pi -Vx(1/a) which I truly don't understand. Many thanks in advance.

 

[Cubist]

I personally think that your calculation and the boxed formula are both correct.

In the solution they do a trick with Vy(a)=pi -Vx(1/a) which I truly don't understand.

I think in their solution they might be trying to use the following...

[MATH] \int_{f(a)}^{f(b)} f^{-1}(y)\,dy + \int_a^b f(x)\,dx=b\cdot f(b)-a\cdot f(a)[/MATH]
See https://en.wikipedia.org/wiki/Integral_of_inverse_functions (click)

however, I'm pretty sure that the following isn't true for all x...

[MATH] \int_{f(a)}^{f(b)} \left(f^{-1}(y)\right)^2\,dy + \int_a^b f(x)^2\,dx = b\cdot f(b)-a\cdot f(a)[/MATH]
it certainly doesn't work if y=x. Therefore, their solution is probably incorrect

 

[Paul Heisenberg]

Thank you for sharing your opinion. I guess I'll have to get in touch with someone that wrote this solution. Thanks.

 

[Cubist]

After re-reading the original question, I realize a mistake that you made. Sorry that I didn't spot this when I wrote post#2. I'll spend some time in the corner ! Anyway, consider the area when α=1000



The relevant area is between the x-axis and the red line. This is going to be a very small area for this α. If we plug it into your equation...

[MATH]V_y(\alpha) = \frac{\alpha\pi}{\alpha+2} \approx 3.1353 \,[/MATH] this volume isn't small and can't represent the rotated area

What you've calculated is the rotation of a different area (to the left of the red line, between the curve and the y axis). To correct your equation, we need to subtract your volume from that of a cylinder with h=1, r=1, volume \(\pi r^2h=\pi \). If you do this, then you'll find that you obtain their answer...

[MATH]\pi - \frac{\alpha\pi}{\alpha+2} = \frac{2\pi}{\alpha+2}[/MATH]
Perhaps sketching a diagram of the area before calculating would help to prevent this kind of error. Tomorrow I'll think again about their method of obtaining this result.

 

[Paul Heisenberg]

Now I see, thank you very much!