Question

Calculus 3: Simplifying \[\int_{0}^{\pi/4} \int_{0}^{\pi/2} \int_{0}^{\sqrt{2}} (\rho\sin\phi \cos\theta)(\rho \sin \phi \sin \theta)\rho^2\sin\phi d\rho d\theta d\phi\]

Answer 1

\[\int_{0}^{\pi/4} \int_{0}^{\pi/2} \int_{0}^{\sqrt{2}} (\rho\sin\phi \cos\theta)(\rho \sin \phi \sin \theta)\rho^2\sin\phi d\rho d\theta d\phi\]
Reducing to what?

Answer 2

first multiply all

Answer 3

\[\int_{0}^{\pi/4} sin^3\phi d\phi \cdot \int_{0}^{\pi/2}\sin\theta\cos\theta \cdot \int_{0}^{\sqrt{2}} \rho^4 d\rho\]

Answer 4

crap this takes a while to write out.

Answer 5

oh its separte integrals like that what?

Answer 6

I'm not really stuck on the process of simplifying it, just the identities that come along with some of the integration, I have forgotten :(

Answer 7

whats your actual question in latex form

Answer 8

okay so

Answer 9

I just ocnfused @radar and @eliassaab , sorry, hahaha

Answer 10

p^4*coss...sin...
root2^5/5*co..sin,,
----------------
and u got sin phi that factors nicely outta the cos and sin theta

Answer 11

so u get some factor *costheta+sintheta

Answer 12

wait no i mean cos theta* sin theta

Answer 13

and cos theta sin theta can be rewritten as

Answer 14

as the sum of addition and difference identities

Answer 15

\[\int_{0}^{\pi/4} sin^3\phi d\phi \cdot \int_{0}^{\pi/2}\sin\theta\cos\theta \cdot \int_{0}^{\sqrt{2}} \rho^4 d\rho\]\[\int_{0}^{\pi/4} (1-cos^2\phi)\sin\phi d\phi \cdot \int_{0}^{\pi/2} \color{red}{sin\theta\cos\theta} \cdot \left.\frac{1}{5}\rho^5\right]_{0}^{\sqrt{2}}\]

Answer 16

sin(a+b) + sin(a-b)
---------------
2

Answer 17

or something like that lets see

Answer 18

ya umm u know how sin theta and cos theta pop out..

Answer 19

That red portion... would it be....

Answer 20

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
sin(A-b)= sin(a)cos(b) - cos(a)sin(b)
sin(a+b)+sin(a-b)= 2 sin(a)cos(b)

Answer 21

Well i know \(sin(2\theta) = 2sin(\theta)cos(\theta)\)

Answer 22

so we can simplify sin(a)cos(b)

Answer 23

with that process

Answer 24

right a-b =0 xD

Answer 25

when a=b

Answer 26

sin(a+b)+sin(a-b)= 2 sin(a)cos(b)
sin(a+a)+sin(a-a)=2sin(a)cos(a)
sin(2a)
------ = sin a cos a
2

Answer 27

okay now just integrate

Answer 28

Oh wait...yeah i was just about to type that. haha

Answer 29

just use a u-sub

Answer 30

For which portion?

Answer 31

hes saying u can use u sub for cos theta sin theta

Answer 32

\[\sin(\theta)\cos(\theta)\]

Answer 33

Oh. Hmm. I was thinking I'd use a u-sub for the more complicated one, the first integral

Answer 34

u=sintheta
du=costheta dtheta so
sin a cos a da = u du

Answer 35

simplifies nicely

Answer 36

but u already know the identity now so

Answer 37

To check your final answer
\[
\int_0^{\frac{\pi }{4}} \left(\int_0^{\frac{\pi
}{2}} \left(\int_0^{\sqrt{2}} \rho \rho \rho
^2 \sin (\theta ) \cos (\theta ) \sin (\phi )
\sin (\phi ) \sin (\phi ) \, d\rho \right) \,
d\theta \right) \, d\phi =\frac{1}{15} \left(4
\sqrt{2}-5\right)
\]

Answer 38

\[\int_0^{\frac{\pi }{4}} \left(\int_0^{\frac{\pi
}{2}} \left(\int_0^{\sqrt{2}} \rho \rho \rho
^2 \sin (\theta ) \cos (\theta ) \sin (\phi )
\sin (\phi ) \sin (\phi ) \, d\rho \right) \,
d\theta \right) \, d\phi =\\\frac{1}{15} \left(4
\sqrt{2}-5\right)
\]

Answer 39

\[\int_{0}^{\pi/4} (1-cos^2\phi)\sin\phi d\phi \cdot \int_{0}^{\pi/2} \color{red}{sin\theta\cos\theta} \cdot \left.\frac{1}{5}\rho^5\right]_{0}^{\sqrt{2}}\]\[\int_{0}^{\pi/4} (1-cos^2\phi)sin \phi d\phi\] If we let u = cos\(\theta\), du = -sin\(\theta\)d\(\theta\)\[-\int_{0}^{\pi/4} (1-u^2)du\] The first part. right?

Answer 40

so final problem is intergral of sinphi ^ 3

Answer 41

yah thats rght

Answer 42

no

Answer 43

change your limits of integration after substitution

Answer 44

Do you have to.... or can you resub before you integrate the limits?

Answer 45

the way you have it written is incorrect

Answer 46

if you want you could write

\[-\int_{u(0)}^{u(\pi/4)} (1-u^2)du\]

Answer 47

\[\int_{0}^{\pi/4} (1-cos^2\phi)\sin\phi d\phi \cdot \int_{0}^{\pi/2} \color{red}{sin\theta\cos\theta} \cdot \left.\frac{1}{5}\rho^5\right]_{0}^{\sqrt{2}} d\theta \]

Answer 48

\[-\int_{1}^{\sqrt{2}/2} (1-u^2)du\]

Answer 49

int cos^2a* sin a da
cos^3a
= -----
3

Answer 50

Ahh.. I see

Answer 51

cos(pi/4)=1/root2

Answer 52

To Check your answer
\[\int_0^{\frac{\pi }{4}} \sin ^3(\phi ) \, d\phi
=\frac{1}{12} \left(8-5 \sqrt{2}\right) \\
\int_0^{\frac{\pi }{2}} \sin (\theta ) \cos
(\theta ) \, d\theta =\frac{1}{2}\\
\int_0^{\sqrt{2}} \rho ^4 \, d\rho =\frac{4
\sqrt{2}}{5}\]

Answer 53

1. \[\left. -u+\frac{1}{3}u^3\right]_{1}^{\sqrt{2}/2}\]2. u= sin\(\theta\), du = \(\cos\theta d\theta\)\[\int_{0}^{1}udu = \left.\frac{1}{2}u^2\right]_{0}^{1}\]3.\[\left.\frac{1}{5}\rho^5\right]_{0}^{\sqrt{2}}\]Haven't simplified them but this is what I have with both of the u-subs.

Answer 54

Oh cool, it matches up with yours. @eliassaab

Answer 55

Solving it I'd get
1. \[\left(-\frac{\sqrt{2}}{2} +\frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3\right)-\left(1+\frac{1}{3}\right)\]

Answer 56

But yeah I get it.... thank you guys!!

Answer 57

and theres a typo there, meant -1*