Multi-variable integration with polar coordination

[MasterWu77]

Homework Statement

$$\int$$$$\int$$ of R ( sin( x^2 + y^2) ) dA where the region 4$$\leq$$ x^2+y^2 $$\leq$$ 49

Homework Equations

not too sure but i know that dy dx = r d(r) d(theta)

The Attempt at a Solution

i don't understand how to change into polar coordinates in order to integrate. I'm unsure of how the 4$$\leq$$ x^2+y^2 $$\leq$$ 49 becomes the r i need to integrate with. and if i do get that then how do i change the other bounds of integrate for the d (theta)

 

[Mark44]

Homework Statement

$$\int$$$$\int$$ of R ( sin( x^2 + y^2) ) dA where the region 4$$\leq$$ x^2+y^2 $$\leq$$ 49

What does "of R" mean in your integral? Does it represent the region you are describing, or is it some factor in the integrand?

Homework Equations

not too sure but i know that dy dx = r d(r) d(theta)

The Attempt at a Solution

i don't understand how to change into polar coordinates in order to integrate. I'm unsure of how the 4$$\leq$$ x^2+y^2 $$\leq$$ 49 becomes the r i need to integrate with. and if i do get that then how do i change the other bounds of integrate for the d (theta)

The region described by the inequalities is an annulus (a shape like a washer) that lies outside the circle x^2 + y^2 = 4 and inside the circle x^2 + y^2 = 49. In this region, r is going to be between two values.

Your integrand is pretty simple to change to polar coordinates, since r^2 = x^2 + y^2.

 

[MasterWu77]

the R refers to the region that is being described. sorry for not clarifying that earlier.

so with r^2 = x^2 + y^2, does that mean the bounds of d(r) would be from 4 to 49? and if so then would the bound of the integral for d(theta) be from 0 to 2(pi) since it is a complete circle?

 

[Mark44]

r does NOT range from 4 to 49. Draw the circles and it should be clear to you why this is so.
$\theta$ does range from 0 to 2$\pi$.

 

[MasterWu77]

ah ok. i believe i see what you mean.

after looking at the circles i think i see that r ranges from 2 to 7 which would be the radius of the circles. and so from there on do i just need to a double integral of sin(r^2) r dr d(theta)? or is there some other little step I'm missing?

 

[Mark44]

That's pretty much it. Your integral should look like this:
$$\int_{\theta = 0}^{2\pi}\int_{r = 2}^7 sin(r^2) r dr d\theta$$

 

[MasterWu77]

ok awesome. i got that integral and then attempted to solve for it.

i used a u substitution for the sin(r^2) where:
u=r^2
du= 2rdr

and i ended up with

-(1/2) $$\int cos(u)d(theta)$$
where the integral is bounded from 0 to 2(pi) and the cos (u) goes from 4 to 49 because of the u substitution i used. i put the answer in and it said that it was incorrect

 

[Mark44]

Your second integral should look like this:
$$(-1/2)\int_{\theta = 0}^{2\pi}\left(cos(49) - cos(4)\right) d\theta$$

Edit: Added multiplier to front of integral.

 

[MasterWu77]

ok i got that but shouldn't there be a (-1/2) out in front of the integral to account for the u substitution?

when i solve for the integral should i get

[(cos(49)-cos(4))* 2pi]

since the there isn't originally a $$\theta$$ so by taking the integral with respect to $$\theta$$ a $$\theta$$ should appear which can be bounded by the 2pi and 0

 

[Mark44]

ok i got that but shouldn't there be a (-1/2) out in front of the integral to account for the u substitution?

Yes, it was in my notes, but I neglected to put that in my post.

when i solve for the integral should i get

[(cos(49)-cos(4))* 2pi]

since the there isn't originally a $$\theta$$ so by taking the integral with respect to $$\theta$$ a $$\theta$$ should appear which can be bounded by the 2pi and 0

Yes, but you need to put in the missing (-1/2). If you evaluate your answer, remember that the numbers are real numbers (i.e., radians), not degrees.

 

[MasterWu77]

ah ok i finally got the right answer! it turns out that the answer is negative but i didn't think that was possible since we're computing an area so wouldn't the answer be a positive number?

 

[Mark44]

No, you're not computing the area of the region R. If you were, the integrand would be 1, not sin(x^2 + y^2). A lot of the surface z = sin(x^2 + y^2) must lie below the x-y plane for the integral to come out negative.

 

[MasterWu77]

o ok that makes sense! thank you very much for helping me with this problem and helping me to understand the concept better