Question

f(x)= (1.6^3)*sqrt(x) and g(x)= x^2. find intersection points.

Answer 1

equate the functions

\[1.6^3\sqrt{x} = x^2\]

divide both sides by root x

\[1.6^3 = \frac{x^2}{x^{1/2}}\]

using index laws

\[1/6^3 = x^{3/2}\]

\[x = (1.6^3)^{2/3}\]

\[x = 1.6^2\]

Answer 2

from the second step: for division with like bases, would subtract 2 - 1/2?

Answer 3

you need to substitute the x value to find y

y = x^2 then y = (1.6^2)^2
y = 1.6^4

the ordered pair is (1.6^2, 1.6^4)

Answer 4

and there is another point of intersection , I just realised... when x=0, then y= 0

so the points of intersection and

(0,0) and (1.6^2, 1.6^4)

Answer 5

what about rotating the region about the y-axis and finding resulting volume using shell method?

Answer 6

I don't know the shell method

Answer 7

\[2pi \int\limits_{a}^{b}(f(x)-g(x))dx\]

Answer 8

all I use for volume of a solid revolved is

\[V = \pi \int\limits_{a}^{b} y^2 dx \]

Answer 9

well the problem is \[V = \pi \int\limits_{0}^{1.6^2} 1.6^3x^{1/2} - x^2 dx\]

rough guess its

\[V = \pi[ 1.6^3 2/3x^3/2 - 1/3x^3]_{0}^{1.6^2}\]

Answer 10

oops

\[V = \pi[1.6^3\frac{2}{3}x^{2/3} - \frac{1}{3}x^3]_{0}^{1.6^2}\]

Answer 11

just never heard it referred to as the shell method

Answer 12

the one you are using is called disc/washer method

Answer 13

ok... thats all I use sorry I think the integral is correct... you just need a multiple of 2 out the front with pi

Answer 14

f(x) is the upper function and g(x) is the lower function.

Answer 15

thank you for your help. :-)

Answer 16

\[2pi \int\limits_{a}^{b}x(f(x)-g(x))dx\]
that is the shell method, sorry. forgot the x.

Answer 17

have a nice day/night.