Expert Answer to "Find the volume of the solid generated by..."

Reeves

Answered question

2021-05-13

Find the volume of the solid generated by revolving the shaded region

stuth1

Skilled2021-05-14Added 97 answers

Your problem was solved

Vasquez

Expert2021-12-24Added 669 answers

The given curve are

$y = 1, x = 0, x = 4 \tan (\frac{π}{3} y)$

The volume of the solid generobed of the shaded are region about y-axis is

$V = \int_{0}^{1} (4 \tan (\frac{π}{3} y))^{2} d y$

$= 16 π \int_{0}^{1} \tan^{2} (\frac{π}{3} y) d y$

$\int n t \frac{π}{3} y = z \Rightarrow \frac{π}{3} d y = d z$

$y = 0 \Rightarrow z = 0 and y = 1 \Rightarrow z = \frac{π}{3}$

$V = 16 π \int_{0}^{\frac{π}{3}} \tan^{2} z (\frac{3}{π}) d z = 48 \int_{0}^{\frac{π}{3}} \tan^{2} z d z$

$= 48 \int_{0}^{\frac{π}{3}} (\sec^{2} z - 1) d z [S i n c e \sec^{2} x - \tan^{2} x = 1]$

$= 48 \int_{0}^{\frac{π}{3}} \sec^{2} z d z - 48 \int_{0}^{\frac{π}{3}} d z$

$= 48 \tan z |_{0}^{\frac{π}{3}} - 48 z |_{0}^{\frac{π}{3}}$

$= 48 (\tan \frac{π}{3} - \tan 0) - 48 (\frac{π}{3} - 0)$

$= 48 (\sqrt{3} - 0) - 48 \frac{π}{3}$

$= 48 [\sqrt{3} - \frac{π}{3}]$

The volume of the solid generated by revolving the shaded region $48 (\sqrt{3} - \frac{π}{3})$

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?