Question

Let R be the region in the first quadrant bounded by the graph of y=8-x^2, the x axis and the y axis

Answer 1

you spin me round round .... i forget the rest of the lyrics tho

Answer 2

this aint a spinner tho is it

Answer 3

No isn't it just indefinite integration? But I still can't come to the correct answer

Answer 4

its an upside down U topped at y=8 ... and shaded about the first quadrant

we want area of volume of rotation? ... area then

Answer 5

wheres it cross the xaxis at? its positive root?

Answer 6

a definite (defined) integral is what you get when you define limits and boundaries. this is going to be bounded from x=0 to x= the intercept

Answer 7

8-x^2 = 0

8 = x^2

sqrt(8) = x

so from 0 to 2sqrt(2)

Answer 8

\[\int_{0}^{2\sqrt{2}}(8-x^2)dx\]
and the rest is basic integration techniques

Answer 9

Why is the top #2 sqrt2 and not 3

Answer 10

because the graph is bounded by the x axis AND 8-x^2

y=0 IS the xaxis soooo

8-x^2 = 0 when x= +- sqrt(8)

Answer 11

8-3^2 aint zero

Answer 12

Oh the. Graph looks like the intercept was 3

Answer 13

lol, mathing it up actually helps ;)

Answer 14

Now if I found the volume after rotating it around the x axis, do I set it up as\[\pi \int\limits_{0}^{?}\]

Answer 15

with 2sqrt2 at the top

Answer 16

you want the area of circles; of which the radius of each is determined by the function

Answer 17

\[\int_{0}^{2\sqrt{2}}\int_{0}^{8-x^2}2\pi r\ dr\ dx\]

Answer 18

Why are there 2 integration signs

Answer 19

because that is one way to work it; do the inside; then the outside :)

go ahead; whats:\[\int_{0}^{8-x^2}2\pi r\ dr\]

Answer 20

That's where I'm lost its a parabola

Answer 21

it is the length of the line from y=0 to y=f(x); the f(x) is defined by the quadratic equation

Answer 22

your adding up all the points between 0 and 8-x^2

Answer 23

Alright let me try this

Answer 24

\[\left(\int_{0}^{2\sqrt{2}}\left[\int_{0}^{8-x^2}2\pi r\ dr\right]dx\right)= \int_{0}^{2\sqrt{2}}\left( \pi (8-x^2)^2-\pi 0^2\right) dx\]
\[\int_{0}^{2\sqrt{2}}\pi (8-x^2)^2\ dx=\pi \int_{0}^{2\sqrt{2}}(84-16x^2+x^4)\ dx\]

Answer 25

the inside says; add up all the points along the way from 0 to f(x) in a circle

the outside says; do that again for every value of x between x=0 and x=2sqrt(2)

so what it ends up doing is adding all the points

Answer 26

I understood the equations except for the top left Hand one

Answer 27

the double integral?

Answer 28

Yeah

Answer 29

worry about your single variable baby calculus stuff at the moment then :) i was just trying to show you something a bit more expansive.

the volume formulas for spheres and circles and what nots come from calculus ...

Answer 30

the area of a circle is simply adding up all the circumferences from r=0 to r=r for instance

Answer 31

And on the curve your adding up every point there is

Answer 32

yep

Answer 33

lets simply use the already know area of a circle; indtead of trying to recreate it :)

pi r^2

\[\int_{0}^{2\sqrt{2}}\pi [f(x)]^2 dx\]