Expert: Paul Klarreich - 1/5/2010

Question
QUESTION: hi

I've some question regarding cal.3

1- how can i calculate the double integral of xcos(2x+y)dA
0<= x <= 2pi/6 , 0<=y<=pi/4

that is what i have done:

xcos(2x+y)dy = xsin(2x+y)
= xsin(2x+pi/4) - xsin(2x)

then how i can integrate dx?

2- Using polar coordinates, how can ievaluate the double integral   sin (x^2+y^2) where R is the region 9<=x^2+y^2<=64 .

i did this:

sin (r2)r dr d%26

where 3<=r<=8

and 0<=%26<=2pi

but i don't get the right answer

3- how i can evaluate this double integral by chaning it to polar integral
(x^2+y^2)^11/2 dxdy , x^2+y^2<=81

your help is very much appericated

ANSWER: Questioner: meme
Country: Qatar
Category: Calculus
Private: No
Subject: cal III
Question: hi

I've some question regarding cal.3

1- how can i calculate the double integral of xcos(2x+y)dA
0<= x <= 2pi/6 , 0<=y<=pi/4

>> You mean  0 <= x <= pi/3 ???

that is what i have done:

xcos(2x+y)dy = xsin(2x+y)
= xsin(2x+pi/4) - xsin(2x)

then how i can integrate dx?

3- how i can evaluate this double integral by chaning it to polar integral
(x^2+y^2)^11/2 dxdy , x^2+y^2<=81

your help is very much appericated
..........................................
{pi/3  {pi/4
|      | x cos(2x + y) dy dx
}0     }0

{pi/3  |y = pi/4
|      | x sin(2x + y) dx
}0     |y = 0

{pi/3
|   ( x sin(2x + pi/4) - x sin 2x ) dx
}0  

looks Ok, now you will have to use integration by parts, I think.

{pi/3
|    x (sin(2x + pi/4) - sin 2x ) dx   <<< factored.
}0  

Now  u = x,  du = dx
    dv = (sin(2x + pi/4) - sin 2x ) dx
    v = -1/2 (cos(...) - cos(2x))

It may be a bit messy, but you should be able to handle it from here.

..............................................

2- Using polar coordinates, how can ievaluate the double integral   sin (x^2+y^2) where R is the region 9<=x^2+y^2<=64 .

I did this:

sin (r2)r dr dt    << using t for theta.  

where 3 <= r <= 8   << looks good.

and 0<= t <=2pi

but i don't get the right answer

.............
Let's try  u = r^2, then du = 2r dr


sin (r2)r dr dt =

sin (u)(du/2) dt

= - 1/2 cos u, from 3 to 8.

= - 1/2 cos(r^2), from 3 to 8.

= - 1/2 (cos 64 - cos 9)


=  -0.65149374615711349676500228484714

Now times 2pi =

-4.0934559335737627030164674850042

Looks weird, but...

------------------------------------------
3- how I can evaluate this double integral by changing it to polar integral
(x^2+y^2)^11/2 dxdy , x^2+y^2<=81

= (r^2)^11/2 r dr

= r^11 r dr

= r^12 dr, for r = 0 to 9,  t = 0 to 2pi.

That should not be too hard.



---------- FOLLOW-UP ----------

QUESTION: thanks alot for your help

regarding to the first question

i can't get the right answer

this is what i've done:

(-pi/6(cos(pi)-cos(4pi/6)))-(1/4*sin(pi))+(1/4sin(4pi/6))

is it wrong?

thankz for your help again

ANSWER: Here is what I did.  Take a look and see if it looks right.

Now  u = x,  du = dx
    dv = (sin(2x + pi/4) - sin 2x ) dx
    v = -1/2 (cos(...) - cos(2x))

uv = x[-1/2 (cos(...) - cos(2x))], from 0 to pi/3

uv = x[1/2 (- cos(...) + cos(2x))], from 0 to pi/3

uv = pi/3[1/2 (- cos(2pi/3 - pi/4) + cos(2pi/3))]

uv = pi/6[(- cos(2pi/3 - pi/4) + cos(2pi/3))]


 {
- | v du =
 }

{
| 1/2 (cos(2x + pi/4) - cos(2x)) dx =
}

1/4 (sin(2x + pi/4) - sin(2x)), from 0 to pi/3

1/4 (sin(2pi/3 + pi/4) - sin(2pi/3)) - 1/4(sin(pi/4)


You have:

(-pi/6(cos(pi)-cos(4pi/6)))-(1/4*sin(pi))+(1/4sin(4pi/6))


---------- FOLLOW-UP ----------

QUESTION: i've emailed the instructor and he told me that i've to do the following:

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

so i did it like this:

(xcos(2x)cos(y)-xsin(2x)sin(y))dy:

xcos(2x)sin(y)+xcos(y)sin(2x)

now how i can integrate dx?

thanks for your help


Answer
QUESTION: i've emailed the instructor and he told me that i've to do the following:

>>  That is an improper use of i've -- the first is OK, but the second should say "i'm to do..." or "I have to do.."  You may abbreviate the first "I have" but not the second.

Anyway, on to some math.

If your instructor says you CAN do this, he is correct.
If your instructor says you MUST do this, he is not correct.

.....................
cos(a+b) = cos a cos b - sin a sinb

so i did it like this:

(x cos(2x) cos(y) - x sin(2x) sin(y)) dy dx:  <<< doing it HIS way.

>> Yes, that is correct.  Then you first integrate w.r.t. y and get:

xcos(2x)sin(y)+xcos(y)sin(2x)   << Looks good.

now how i can integrate dx?

You still have to use I.B.P. on this.  Now you have two terms to handle.  Of course, the  sin y  and cos y things are considered constants.

First:
xcos(2x),  let  u = x, du = dx,   dv = cos(2x)dx, v = 1/2 sin(2x)

Get:

x/2 sin(2x) - INT 1/2 sin (2x) dx

x/2 sin(2x) + 1/4 cos (2x)  <<< PART 1, times  sin y

Second:
x sin(2x), let u = x,  du = dx,  dv = sin(2x)dx,  v = -1/2 cos(2x)

Get:

-x/2 cos(2x) - INT (-1/2 cos(2x))

-x/2 cos(2x) + 1/4 sin(2x)  <<< Part 2,  times cos y

(x/2 sin(2x) + 1/4 cos (2x)) sin y + (-x/2 cos(2x) + 1/4 sin(2x)) cos y

x/2[ sin(2x) sin y - cos(2x) cos y] + 1/4[cos 2x sin y + sin 2x cos y]

= - x/2 cos(2x + y) + 1/4 sin(2x + y)

Now you can put your y values(0 to pi/4) and your x-values.

==============================
Doint it all at once:

x cos(2x+y) dy first:    <<< doing it MY way.

x sin(2x + y)   << integral of cos(... + y) =

let  u = x, du = dx,  dv = sin(2x + y) dx,  v = - 1/2 cos(2x + y)

Get:

-x/2 cos(2x + y) - INT (- 1/2 cos(2x + y)) dx

-x/2 cos(2x + y) + 1/4 sin(2x + y)

AND THAT LOOKS THE SAME AS BEFORE.  GIVE IT ANOTHER SHOT; as above:

Put your y values(0 to pi/4) and your x-values( 0 to pi/3?).

Your teacher is a little inflexible.  (Well, he is probably young.  I might have some stronger language to describe him, but I am really a nice guy.)