Question

Find the volume of the solid formed by rotating the region inside the first quadrant enclosed by
y=x^3,y=16x
about the x-axis.

Answer 1

your boundaries are
\(y=x^3\) and \(y=16x\)

and it is rotated about the x,axis correct?

Answer 2

x-axis *

Answer 3

yes

Answer 4

that will look like a washer method

Answer 5

do you know the washer method?

Answer 6

(don't be ahsamed not to know, if you don't know it is ok. the only person i can not help is the silent one)

Answer 7

i do but have trouble with the set up

Answer 8

ok, lets figure the limits of integration together

I will name them differently for our convinience
\(f(x)=16x\)
\(g(x)=x^3\)

Answer 9

lets find the points where these two functions intersect

Answer 10

16x=x³
16=x²

(and disregard negative x solutions because we are only in the 1st quadrant)

Answer 11

x=?

Answer 12

4

Answer 13

yes, and one more?

Answer 14

0

Answer 15

yes

Answer 16

So our limits of integration (for the region bounded by 16x and x³) is (x=0 and x=4) \
(i.e. the intersection points after which the region stops)

Answer 17

So far we know \(\large\color{slate}{\displaystyle\int\limits_{0}^{4}}\)

Answer 18

okay got it and hw to determine inner vs outer for the formula

Answer 19

yes, and if you were to graph the two functions
\(f(x)=16x\) and \(g(x)=x^3\)
then for all values on the interval [0,4] (which is the interval we need)
it is true that f(x)≥g(x)

Answer 20

So, g(x) is always below the f(x) in our case. right?

Answer 21

(you can test it using any value between 0 and 4)

Answer 22

okay great thanks , makes more sense now , you were a big help!!

Answer 23

you are thinking me already, we aren;t done yet, aren we?

Answer 24

thanking*