Rotational volume

[fred2028]

The question reads:

Find the volume of the solid given by the area bound by y = tanx, y = 0, and x = pi/4 being rotated along the y = -1 axis.

So, here's what I did. I moved up the entire thing so that it is rotated about the x-axis. That means the upper function is y = tanx + 1, and the lower function is y = 1, all this being rotated about x-axis. To find volume, I did:

integral_from_0_to_pi/4(pi * (tanx + 1 - 1)^2)dx
= pi * integral_from_0_to_pi/4((tanx)^2)dx
= pi * integral_from_0_to_pi/4((secx)^2 - 1)dx
= pi * [tanx - x] from 0 to pi/4
= pi * (1 - pi/4)
= Some decimal number beginning with 0.6 ....

The back of the book gives an answer that evaluates to about 2.8. What did I do wrong?

 

[galactus]

Shells may be OK here instead of washers. You have a +1-1 in your work. That is just 0.

$\displaystyle 2{\pi}\int_{0}^{tan(\frac{\pi}{4})}(1+y)(\frac{\pi}{4}-tan^{-1}(y))dy$

Here's washers:

$\displaystyle {\pi}\int_{0}^{\frac{\pi}{4}}[(tan(x)+1)^{2}-1]dx={\pi}+{\pi}ln(2)-\frac{{\pi}^{2}}{4}\approx 2.851777644$


Here is a graph to help visualize:

 

[BigGlenntheHeavy]

Not trying to steal galactus's thunder. but here is washers expanded my way:

$\displaystyle \pi \int_{0}^{\frac{\pi}{4}}[(tan(x)+1)^{2}-(0+1)^{2}]dx$

$\displaystyle \pi \int_{0}^{\frac{\pi}{4}}[tan^{2}(x)+2tan(x)]dx$

$\displaystyle \pi \int_{0}^{\frac{\pi}{4}}tan^{2}(x)dx+2\pi \int_{0}^{\frac{\pi}{4}}tan(x)dx$

$\displaystyle \pi \int_{0}^{\frac{\pi}{4}}[sec^{2}(x)-1]dx-2\pi ln|cos(x)|]_{0}^{\frac{\pi}{4}}$

$\displaystyle \pi(tan(x)-x)]_{0}^{\frac{\pi}{4}}-2\pi[ln(\frac{1}{\sqrt2})-ln(1)]$

\(\displaystyle \pi(1-\frac{\pi}{4})-2\pi({ln(1)-\frac{ln(2)}{2}) = \pi-\frac{\pi^{2}}{4}+\pi ln(2) = 2.85177764362\)

 

[fred2028]

Wow, those answers are really detailed.

Shouldn't the integrand be just radius squared? And isn't the radius given by [(tanx + 1) - 1]? From what I see here, the correct integral is actually the volume of the larger solid (tan) minus the volume of the inner solid (cylinder).