Question

Integration using Washer method

Answer 1

What Im doing:

\[\int\limits_{0}^{1}\pi(3\sqrt{x}^2+\int\limits_{1}^{4}\pi(3-x)^2\]

Answer 2

@agent0smith

Answer 3

You split the region into two integrals, good good good.

Dealing with the second region,
Your outer radius R=4-x, ya?
with inner radius r=1.

So then, dV=pi(R^2-r^2)dx

Answer 4

wouldnt you add them?

Answer 5

Add the integrals like you've done? Yes.

I'm just worried about the stuff inside of the second integral.
\(\large\rm \pi\left[(4-x)^2-1^2\right]\ne\pi(3-x)^2\)

Answer 6

ohhhh

Answer 7

Oh I just noticed, your first integral looks a little off as well.
You didn't subtract the inner radius in that one.

Answer 8

I forgot the functions need to be inside parentheses

Answer 9

Oh, yea. I had it on my work but it was a little messy so accidentally left it off lol

Answer 10

R=3sqrt(x)
r=1

\(\large\rm dV_1=\pi\left[(3\sqrt{x})^2-1^2\right]dx\)

Answer 11

Ah :D

Answer 12

\[\int\limits_{0}^{1}\pi[(3\sqrt{x})^2-(1)^2]+\int\limits_{1}^{4}\pi[(4-x)^2-1^2] = dV _{1+2}\]

Answer 13

That look better?

Answer 14

\(\large\rm =\int dV_1+\int dV_2\)
ya looks better besides that last equals sign :)

Answer 15

lol just trying to keep the notation flowing

Answer 16

I'll let you know what I come up with

Answer 17

When you integrate (add up) all of those dV1's you could call it V1,
so we end up with
\(\large\rm =V_{1+2}\)

Answer 18

k cool

Answer 19

Those limits aren't right.

Answer 20

No? :O
Hmmmm they look right to me.
I better check again.

Answer 21

\(\large\rm -x+4=3\sqrt x\)
squaring,
\(\large\rm x^2-8x+16=9x\)
subtracting 9x,
\(\large\rm x^2-17x+16=0\)
factoring
\(\large\rm (x-1)(x-16)=0\)

So we get x=1 and x=16,
but it turns out x=16 is extraneous, ya?

Answer 22

That limit is fine, but the lower of the first integral and upper of the second are not. The region is bounded by \(y=1\), not \(y=0\).

Answer 23

Should be \(\displaystyle\int_{1/9}^1\cdots+\int_1^3\cdots\). The setup is fine otherwise.

Answer 24

Oh the leftmost boundary is NOT x=0 is it?
woops, I didn't notice that D:

Answer 25

the second of the upper was the x intercept

Answer 26

\(\large\rm 1=3\sqrt x\)

Answer 27

of y=4-x that is

Answer 28

And for the rightmost boundary
\(\large\rm 4-x=1\)

Answer 29

ah I see

Answer 30

good catch @HolsterEmission

Answer 31

I've got to go to lecture now unfortunately. Ill finish this up afterwards and post my results

Answer 32

UPDATE: Ok, could could someone help me understand the bounds here?

Answer 33

I want to flip the purple shaded region right?

Answer 34

Here's a sketch of the region of interest: