Question

Now Back to CALCULUS 3 !

Answer 1

\[2/\pi \int\limits_{0}^{1} \int\limits_{0}^{\pi/2} \sin(u)(v+cosu)^1/2 du dv\]

Answer 2

@ayeshaafzal221

Answer 3

Team work New title Rookie. lol what's that. yay!!!.

Answer 4

Is that supposed to be \(\sqrt{v+\cos(u)}\)?

Answer 5

@ yes

Answer 6

is a problem under average value

Answer 7

@SolomonZelman hint?

Answer 8

since the region is rectangular region we can change the order of integration dvdu i think it will be much easier to integrate.

Answer 9

when integrating with respect to v, u is constant, and vice versa

Answer 10

How did we get "u" and "v"? Have we already transformed things a bit?

Try \(w = v+\cos(u)\)

Answer 11

Latex Suggestion:
When you write the power \(^{1/2}\), type `^{1/2}`

Answer 12

\[du=dw/-\sin(u)\] @tkhunny

Answer 13

Also, space is `~`

\(\color{#000000 }{\displaystyle 2/\pi \int\limits_{0}^{1} \int\limits_{0}^{\pi/2} \sin(u)(v+\cos u)^{1/2}~du~dv}\)

Answer 14

Oh nice.

Answer 15

kk

Answer 16

ok, now do the sub tkhunny gave you

Answer 17

And you should get (when integrating with respect to \(u\));

\(\color{#000000 }{\displaystyle w=v+\cos(u)}\)
\(\color{#000000 }{\displaystyle -dw=\sin(u)~du}\)

Answer 18

@tkhunny nice one. i figured out the first integral

Answer 19

@SolomonZelman yes I did w=v+cos (u)

Answer 20

\[2/3\int\limits_{?}^{?} (v+cosu)^3/2\]

Answer 21

Rule of Thumb: ALWAYS try what's under the radical - especially in a textbook problem.

Answer 22

how can I tackle dv

Answer 23

Didn't learn that LaTex format, I see.

Answer 24

And I have pluged in my values for u \[2/3(v+\cos \pi/2)^3/2-2/3(v+1)^2/3\]

Answer 25

\(\color{#000000 }{\displaystyle 2/\pi \int\limits_{0}^{1}\frac{-2}{3}\left[ v^{3/2}-(v+1)^{3/2}\right]~dv}\)

\(\color{#000000 }{\displaystyle \frac{4}{3\pi} \int\limits_{0}^{1}\left[ (v+1)^{3/2}-v^{3/2}\right]~dv}\)

Answer 26

this is what I got for du,

and then you can integrate term by term...

Answer 27

\[\int\limits_{0}^{1} thats function ~~~~~~dv\]

Answer 28

Ok let me try

Answer 29

Is it different from mine /

Answer 30

\(\color{#000000 }{\displaystyle \frac{4}{3\pi} \int\limits_{0}^{1}\left[ (v+1)^{3/2}-v^{3/2}\right]~dv}\)

\(\color{#000000 }{\displaystyle \frac{4}{3\pi} \int\limits_{0}^{1} (v+1)^{3/2}dv-\frac{4}{3\pi} \int\limits_{0}^{1} v^{3/2}dv}\)

Answer 31

what is different from what? ....

Answer 32

How did you get 4/3pi

Answer 33

instead I have \[-2/3~~~[v^3/2-(v+1)^3/2]\]

Answer 34

\(\color{#000000 }{\displaystyle 2/\pi\int\limits_{0}^{1} \color{#0000ff}{ \int\limits_{0}^{\pi/2} \sin u(v+\cos u)^{1/2}~du}~dv}\)

\(\color{#0000ff }{\displaystyle w=v+\cos u}\)
\(\color{#0000ff }{\displaystyle dw=-\sin u~du\quad \Longrightarrow\quad -dw=\sin u~du}\)

(I will substitute the initial variable (u), later, instead of changing the limits of integration)

\(\color{#000000 }{\displaystyle 2/\pi\int\limits_{0}^{1} \color{#0000ff}{ \int\limits_{0}^{\pi/2} -(w)^{1/2}~dw}~dv}\)

\(\color{#000000 }{\displaystyle 2/\pi\int\limits_{0}^{1} \color{#0000ff}{ -\left[\frac{2}{3}w^{3/2}\right]_{u=0}^{u=\pi/2}}~dv}\)

\(\color{#000000 }{\displaystyle 2/\pi\int\limits_{0}^{1} \color{#0000ff}{ -\left[\frac{2}{3}(v+\cos(\pi/2))^{3/2}-\frac{2}{3}(v+\cos(0) )^{3/2}\right]}~dv}\)

\(\color{#000000 }{\displaystyle 2/\pi\int\limits_{0}^{1} \color{#0000ff}{ \left[\frac{2}{3}(v+\cos(0) )^{3/2}-\frac{2}{3}(v+\cos(\pi/2))^{3/2}\right]}~dv}\)

\(\color{#000000 }{\displaystyle 2/\pi\int\limits_{0}^{1} \color{#0000ff}{ \left[\frac{2}{3}(v+1 )^{3/2}-\frac{2}{3}(v+0)^{3/2}\right]}~dv}\)

\(\color{#000000 }{\displaystyle \frac{4}{3\pi}\int\limits_{0}^{1} \left[(v+1 )^{3/2}-v^{3/2}\right]~dv}\)

That is how I integrated that du, and notice the \(\color{red}{\displaystyle \frac{4}{3\pi}}\) in there ...

Answer 35

Then, as I pointed out,
you integrate term by term.

\(\color{#000000 }{\displaystyle \frac{4}{3\pi} \int\limits_{0}^{1} (v+1)^{3/2}dv-\frac{4}{3\pi} \int\limits_{0}^{1} v^{3/2}dv}\)

Answer 36

I think what I have for du is right

Answer 37

look through what I did and tell me where I made a mistake ...

Answer 38

LOL multiply mine by 2pi/3 and you get the same thing as yours. I multiply the constants at the end. Oh great. Nice one. I will integrate by part now.

Answer 39

you don't mean uv-\(\int\)vdu?

Answer 40

the 2/pi before the integral.

Answer 41

No no no not by part.

Answer 42

ok, tell me the numerical (exact value) of the integral after you integrate dv, and evaluate at the limits of integration...

Answer 43

ummm well I got approx 9.3084

Answer 44

Try again?

Answer 45

What I get is:

\(\color{#000000 }{ \displaystyle \frac{4}{3\pi} \cdot \frac{2}{5}\cdot (2)^{5/2}-\frac{4}{3\pi} \cdot \frac{2}{5}\cdot (1)^{5/2} -\frac{4}{3\pi} \cdot \frac{2}{5}\cdot (1)^{5/2}+\frac{4}{3\pi} \cdot \frac{2}{5}\cdot (0)^{5/2}}\)

\(\color{#000000 }{ \displaystyle \frac{8}{15\pi} \cdot (2)^{5/2}- \frac{16}{15\pi}\cdot (1)^{5/2} }\)

\(\color{#000000 }{ \displaystyle \frac{32\sqrt{2}}{15\pi} - \frac{16}{15\pi}}\)

\(\color{#000000 }{ \displaystyle \frac{32\sqrt{~2~}-16}{15\pi}}\)

Answer 46

just adding, that this is of course less than what you got...

Answer 47

verification that I am correct:

http://www.wolframalpha.com/input/?i=%282%2F%CF%80%29+integral+sin%28x%29%28z%2Bcos%28x%29%29%5E%281%2F2%29+dx+from+x%3D0+to+x%3D%CF%80%2F2

http://www.wolframalpha.com/input/?i=%284%2F%283%CF%80%29%29+integral+%28x%2B1%29%5E%283%2F2%29-x%5E%283%2F2%29+dx+from+x%3D0+to+x%3D1

Answer 48

yeah. I'll go over it again. thank you so so so so so so so so much @SolomonZelman

Answer 49

No problem ...

Answer 50

Gracias:-

Answer 51

@SolomonZelman Please could you show me how you got \[32\sqrt{2}\]

Answer 52

from \[(2)^5/2\\]

Answer 53

\(8\cdot 2^{5/2}\), actually.

Answer 54

yes that

Answer 55

\(8\cdot 2^{5/2}=8\cdot2^{4/2~+~1/2}=8\cdot 2^{4/2}\cdot 2^{1/2}=8\cdot 4 \cdot \sqrt{2}=32\sqrt{2}\)

Answer 56

that's just algebra ...

Answer 57

I have to go for about an hour right now... hope that you have clarified everything about the problem....

Answer 58

Oh yea sure. thanks.