Question

Calc III students! Can you give me some difficult integral problems you may have been assigned without answers? (I need to study for the GRE)

Answer 1

What is in the syllabus?

Answer 2

..? There is no syllabus?

Answer 3

I mean what are the topics in Calc III?

Answer 4

Uh essentially, triple integrals, polar, spherical, vector integration, partial derivs. That type of stuff, Idk, really anymore, hence the question. It's been like 4 years since I had it

Answer 5

\[\int_{-\infty}^{\infty}e^{-x^2}dx\]

Answer 6

thats a fun integral :)
I think @eliassaab sir has some website to practice calcIII problems

Answer 7

...Don't you have to go into the complex plane for that? It's the definition of the zeta function or something right?

Answer 8

straight computation yields a divergent integral to infinity, no?

Answer 9

no, hint
\(\int_a^b f(x) dx = \int_a^bf(y)dy\\x^2+y^2=r^2\)

Answer 10

oh god... I have to go to into spherical coordinates?...It really has been a while. I have no idea what your hint means if that is not it

Answer 11

no, polar

Answer 12

okhm

Answer 13

\[\int\limits 1+\sin \theta divcos \theta \sqrt{4\cos^2\theta+(1+\sin \theta)^2}d\]

Answer 14

\[\int_{-\infty}^{\infty}e^{-x^2}dx\]\[x^2=r^2-y^2\]\[-\infty <y^2<\infty\] \[-\infty <r^2<\infty\]\[\int \int e^{-(r^2-(rsin\theta^2)}rdrd\theta\]

Answer 15

the squared sould be on the entire rsintheta term, what do you think?

Answer 16

@danish071996 I've never seen divcos, what function is that?

Answer 17

oh it is division

Answer 18

\[\int\limits 1+\frac{\sin \theta}{ cos \theta} \sqrt{4\cos^2\theta+(1+\sin \theta)^2}d\theta\] Is this it?

Answer 19

How do we know that e^-x^2=e^-y^2?

Answer 20

well, the itegral of those

Answer 21

no only 1+sin theta in the numerator

Answer 22

ah \[\int\limits \frac{1+\sin \theta }{cos \theta \sqrt{4\cos^2\theta+(1+\sin \theta)^2}}d\theta\]

Answer 23

\(\int_{-\infty}^{\infty}e^{-x^2}dx=\int_{-\infty}^{\infty}e^{-y^2}dy\\
\int_{-\infty}^{\infty}e^{-x^2}dx*\int_{-\infty}^{\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2}e^{-x^2}dx \ dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx = \\ \int_0^{2\pi}\int_0^\infty e^{-r}r\ dr \ d\theta = \int_0^{2\pi}\frac{1}{2} d\theta = \pi \)

so \( (\int_{-\infty}^\infty e^{-x^2}dx)^2=\pi\) thus \[\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}\]

Answer 24

@zzr0ck3r , can you explain the initial reasoning of f(x)=f(y)?

Answer 25

\(\int_2^3x^2dx=\int_2^3y^2dy\)

Answer 26

oh, just a change of variable? Bt why do it?

Answer 27

well keep reading and it should become clear.

Answer 28

unfortunately it cuts off, also wouldn't we have to divide by the integral at the end or take the square root since we are essentially squaring it?

Answer 29

\(\int_{-\infty}^{\infty}e^{-x^2}dx=\int_{-\infty}^{\infty}e^{-y^2}dy\\
\implies (\int_{-\infty}^{\infty}e^{-x^2}dx)(\int_{-\infty}^{\infty}e^{-y^2}dy)= (\int_{-\infty}^{\infty}e^{-x^2}dx)^2\)

Answer 30

I see that it works quite beautifully into polar though

Answer 31

but, yea, that last post is where I do not follow, Is there something in the transition that negates the exponent?

Answer 32

\(\int_{-\infty}^{\infty}e^{-x^2}dx=\int_{-\infty}^{\infty}e^{-y^2}dy\\
\int_{-\infty}^{\infty}e^{-x^2}dx*\int_{-\infty}^{\infty}e^{-y^2}dy= \\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2}e^{-x^2}dx \ dy=\\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dx = \\ \int_0^{2\pi}\int_0^\infty e^{-r}r\ dr \ d\theta = \\ \int_0^{2\pi}\frac{1}{2} d\theta = \pi

\)

Answer 33

So what we have shown is

\((\int_{-\infty}^\infty e^{-x^2}dx)^2= \int_{-\infty}^{\infty}e^{-x^2}dx*\int_{-\infty}^{\infty}e^{-y^2}dy=\pi\)

so

\(\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}\)

Answer 34

oh, ohk, I will need to repeat this

Answer 35

Where did you find that problem?

Answer 36

On here I think.

Answer 37

hmm, ok, it's a really good problem to show me I need to review a lot. I would have never thought of that

Answer 38

Thank you, I really appreciate it!

Answer 39

not many people do, don't worry. It is a very clever trick and I am sure most people that know it have been taught it and did not just come up with it.

Answer 40

If you have any others you stumble upon, please post more if you have them. (A few grad students have said the Math GRE is just the hardest calc III integrals essentially, so I want to go through a lot of them in order to avoid a surprise)

Answer 41

https://www.youtube.com/watch?v=fWOGfzC3IeY

Answer 42

Use firefox to access my site on
http://moltest.missouri.edu/mucgi-bin/calculus.cgi

Answer 43

That is amazing, thank you elias!

Answer 44

it also shows a complete solution for each problem
http://gyazo.com/7e02f0160fb6efc51a357fca070dfc21

Answer 45

nice!

Answer 46

YW @FibonacciChick666