Question

Find the volume of the solid generated by revolving the described region about the given axis:

The region enclosed by the curve y = sqrt(5x) , the lines y = 2x-5 and x = 0, rotated about the -axis.

Answer 1

which axis are you rotating around?

Answer 2

y-axis, sorry.

Answer 3

first equate 2x-5=sqrt(5x)
square on both sides and simplify and solve for x use the value of x to find y
the rearrange curve to give x=y^2/5
then integrate \[2\pi \int\limits_{0}^{y} (y ^{2}/5)^{2}dy\] where y (upper limit) is the value found above

Answer 4

hi!

Answer 5

do u want me to TEACH YOU THIS STUFF

Answer 6

no you are supposed to find volume of solid given by the curve between line y=2x-5 and line x=o as it is rotated around y-axis so the curve is rearranged to find x in terms of y and used in the integral and limits are defined along y-axis

Answer 7

@dan815 I can give you notes :)

Answer 8

Use method of cylindrical shells \[\large V = \int\limits_{a}^{b}2 \pi\cdot x\cdot f(x)dx\]\[\large f(x) = \sqrt{5x} - (2x-5) = \sqrt{5x}-2x+5\]\[\large V =2 \pi \int\limits_{0}^{5}x(\sqrt{5x}-2x+5)dx\]