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.

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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.

Calculus1
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your boundaries are \(y=x^3\) and \(y=16x\) and it is rotated about the x,axis correct?
x-axis *
yes

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that will look like a washer method|dw:1437943945723:dw|
do you know the washer method?
(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)
i do but have trouble with the set up
ok, lets figure the limits of integration together I will name them differently for our convinience \(f(x)=16x\) \(g(x)=x^3\)
lets find the points where these two functions intersect
16x=x³ 16=x² (and disregard negative x solutions because we are only in the 1st quadrant)
x=?
4
yes, and one more?
0
yes
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)
So far we know \(\large\color{slate}{\displaystyle\int\limits_{0}^{4}}\)
okay got it and hw to determine inner vs outer for the formula
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)
So, g(x) is always below the f(x) in our case. right?
(you can test it using any value between 0 and 4)
okay great thanks , makes more sense now , you were a big help!!
you are thinking me already, we aren;t done yet, aren we?
thanking*
i just struggle with the set up I just worked out that practice problem and solved it form there
ok, but, just to make sure can you show me your integral that you will set up?

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