Question

Calculus help!

Answer 1

whats the question?

Answer 2

i just attached it

Answer 3

is there anyway of solving without trig function?

Answer 4

i doubt it

Answer 5

so i made it this far
\[9\pi \int\limits_{0}^{\sqrt3} \frac{ 1 }{ 9+x^2 }dx\]

Answer 6

@TheSmartOne

Answer 7

Use the arc tan method, to integrate?

Answer 8

heres the enclosed solid where the green line is \[\sqrt{3}\]

Answer 9

Is there any other way? @iTz_Sid

Answer 10

Honestly. Thats the only way I see of solving it. :/

Answer 11

@zepdrix

Answer 12

@Astrophysics

Answer 13

Why do you need another way? :o
Trig sub works so nicely.

Answer 14

\[\large\rm x=3\tan \theta\]

Answer 15

teacher hasn't taught trig sub yet so I'm trying to see if there is another way

Answer 16

is there another way or how does trig sub Work?

Answer 17

I'm not sure of another way, sorry :(
\[\large\rm 9\pi \int\limits\frac{1}{9+x^2}dx\]The idea is to make use of your Pythagorean Identity to remove the addition in the problem.\[\large\rm x=3\tan \theta\]Our integral becomes,\[\large\rm 9\pi \int\limits\limits\frac{1}{9+9\tan^2\theta}dx\]Factoring out a 1/9 gives us,\[\large\rm \pi\int\limits\frac{1}{1+\tan^2\theta}dx\]Notice from this point you can apply your Pythagorean Identity, ya?\[\large\rm 1+\tan^2\theta=\sec^2\theta\]

Answer 18

\[\large\rm x=3\tan \theta\]Of course we need to replace the differential as well,\[\large\rm dx=3\sec^2\theta~d \theta\]

Answer 19

\[\large\rm \pi\int\limits\limits\frac{1}{1+\tan^2\theta}3\sec^2\theta~d \theta\]

Answer 20

Lil confusing? :o

Answer 21

I'm not sure why you would be assigned a problem that makes use of a concept you haven't learned yet :\

Going ahead in the book or something?

Answer 22

a little lol

Answer 23

no but yeah he left it as a homework problem, he expects us to do our research lol

Answer 24

Ooo.. dat.. man :[

Answer 25

so can we replace the\[1+\tan ^{2}\theta \] now

Answer 26

Maybe we back up a tiny sec.
Because it might not be totally clear why I chose 3tangent for the substitution.

Answer 27

yeah that would be a big help

Answer 28

\[\large\rm 9\pi \int\limits\limits\frac{1}{9+x^2}dx\]If we pull a 9 out of each term in the denominator, right from the start,\[\large\rm \frac{9\pi}9 \int\limits\limits\limits\frac{1}{1+\frac{x^2}{9}}dx\]Let's bring this 9 into the square with the x though,\[\large\rm \pi\int\limits\limits\limits\limits\frac{1}{1+\left(\frac{x}{3}\right)^2}dx\]

Our denominator now has the form \(\large\rm 1+(stuff)^2\)

Answer 29

This is very helpful, because we know that our Pythagorean Identity has the form \(\large\rm 1+(\tan\theta)^2\)

So we just replace the "stuff" with tangent and we're good to go.

Answer 30

\[\large\rm \frac x3=\tan \theta\]

Answer 31

Multiplying by 3 gives us \(\large\rm x=3\tan\theta\)

I guess before, I was thinking about it a little differently.
If we can somehow get a 9 on each term, then factor it out,
then we have effectively created the 1+tan^2.

Answer 32

I think that second way of thinking is a little less algebraically complicated...
but I dunno.

Answer 33

\[\large\rm \pi\int\limits\limits\limits\limits\limits\frac{1}{1+\left(\color{royalblue}{\frac{x}{3}}\right)^2}\color{orangered}{dx}\]
Our substitution looks like this:\[\large\rm \color{royalblue}{\frac x3=\tan \theta}\]Taking derivative to find our differential replacement,\[\large\rm \frac{dx}3=\sec^2\theta~d \theta\qquad\to\qquad \color{orangered}{dx=3\sec^2\theta~d \theta}\]

Answer 34

\[\large\rm \pi\int\limits\limits\limits\limits\limits\limits\frac{1}{1+\left(\color{royalblue}{\frac{x}{3}}\right)^2}\color{orangered}{dx}\quad=\quad \pi\int\limits\limits\limits\limits\limits\limits\frac{1}{1+\left(\color{royalblue}{\tan \theta}\right)^2}\color{orangered}{3\sec^2\theta~d \theta}\]

Answer 35

Sorry I'll take a breath lol

Answer 36

yeah know i see what how you got it

Answer 37

lol

Answer 38

\[\large\rm 3\pi\int\limits\frac{\sec^2\theta}{1+\tan^2\theta}d \theta\]Applying our Pythagorean Identity in the denominator (that was the whole purpose of doing this goofy method),\[\large\rm 3\pi\int\limits\frac{\sec^2\theta}{\sec^2\theta}d \theta\]

Answer 39

\[\large\rm 3\pi\int\limits d \theta\]The actual integration step is pretty trivial, ya? :o
It's all the stuff surrounding it that makes it tricky.

Answer 40

Yeah it is a bit tricky

Answer 41

Hmm ok let's at least try to finish this one up :D
Maybe you can look back at it later and make sense of it.

Answer 42

So integrating 1 d(theta) gives us theta.\[\large\rm 3\pi\int\limits\limits d \theta\quad=\quad 3\pi \theta\]In the same way that integrating 1*dx would give us x.
Hmm but what is this theta, it's not inside of any trig function.
We'll have to go back to our substitution to make sense of it.\[\large\rm \frac x3=\tan \theta\]We need to solve for theta,\[\large\rm \theta=\arctan\left(\frac x3\right)\]Hopefully that step makes sense, yes? :o

Answer 43

yes it does

Answer 44

\[\large\rm 3\pi\int\limits\limits\limits d \theta\quad=\quad 3\pi \theta\]\[\large\rm 3\pi\int\limits\limits\limits d \theta\quad=\quad 3\pi \arctan\left(\frac x3\right)\]So instead of looking for new boundary values for theta,
we're just undoing our substitution and using the old boundaries.

Answer 45

\[\large\rm 3\pi \arctan\left(\frac x3\right)|_0^{\sqrt3}\]Giving us,\[\large\rm 3\pi \left[\arctan\left(\frac{\sqrt3}{3}\right)-\arctan(0)\right]\]

Answer 46

Your a lifesaver!

Answer 47

Thanks it makes sense too!

Answer 48

You remember how to deal with that arctangent type stuff?
How to turn it into an angle?

Answer 49

not quite

Answer 50

\[\large\rm \arctan\left(\frac{\sqrt3}{3}\right)=\phi\qquad\to\qquad \tan \phi=\frac{\sqrt3}{3}\]I guess you have to remember a little bit of your unit circle.
This is the smaller angle, ya?
\(\large\rm \phi=\frac{\pi}{6}\) I think.

Answer 51

And tan0=0. Therefore arctan(0)=0.

Answer 52

\[\large\rm 3\pi \left[\arctan\left(\frac{\sqrt3}{3}\right)-\arctan(0)\right]\]becomes,\[\large\rm 3\pi \left[\frac{\pi}{6}-0\right]\]

Answer 53

yeah i must review my trig functions

Answer 54

And don't forget, Wolfram is a great tool for checking your work! :)
https://www.wolframalpha.com/input/?i=integral+from+0+to+sqrt3+of+9pi%2F(9%2Bx%5E2)dx

Answer 55

Thanks!