Question

Evaluate the double integral (5x-8y)dA D is bounded by the circle with center the origin and radius 5

Answer 1

\[\large \int_R 5x-8y \quad dA\]

\[\large R: x^2 + y^2 = 25\]

gut reaction is that you have to pound it out in Cartesian.

Answer 2

@ganeshie8 @freckles @Empty

Answer 3

Why not just convert to polar right away?
\[\begin{cases}x=r\cos\theta\\y=r\sin\theta\\dA=dx\,dy=r\,dr\,d\theta\end{cases}\]so
\[\begin{align*}\iint_D(5x-8y)\,dA&=\int_0^{2\pi}\int_0^5(5r\cos\theta-8r\sin\theta)r\,dr\,d\theta\\[1ex]&=\left(\int_0^{2\pi}(5\cos\theta-8\sin\theta)\,d\theta\right)\left(\int_0^5r^2\,dr\right)\end{align*}\]
Meanwhile, via Cartesian coordinates you would have to compute
\[\begin{align*}\iint_D(5x-8y)\,dA&=\int_{-5}^5\int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}(5x-8y)\,dy\,dx\end{align*}\]which is doable, but I feel it's more work than needed.

Answer 4

Alternatively, since \(5x\) and \(8y\) are continuous, you can try writing this as a line integral by applying Green's theorem, but again, more work than necessary.

Answer 5

Before diving in, it's also a good idea to just think about what the integral will look like. Here's something we can do mentally:

\[5\iint xdA-8 \iint ydA \]

Just look at the form of one of these integrals:

\[\iint ydA \]

Each of these halves cancel each other out because the region of integration is symmetric and we have an odd function, so without evaluating anything, it turns out the entire integral is zero.

That's good, cause the thing in there did look kind of like the curl of a vector, so we could have turned this into a line integral \[6x-8y=\nabla \times F\]
\[\iint 6x-8y dA = \iint \nabla \times F dA = \oint F \cdot dS= 0 \]

So that's cool, you don't even have to solve for F to know it's a conservative field, neato. :P

Answer 6

an aside , perhaps an unneresting one, but Mary Boas' book is big on going into params late...it can lead to incredibly concise solutions compared to the often long long stuff you find on Paul's Online math notes (as brilliantly useful as those notes are). Boas often ends up in trig territory not by switching to polar co-ords or whatever but by starting out in Cartesian and then maybe bringing in a trig sub to do an integral.

here you don't even have to do that....


\(\large \int_{x=-5}^5\int_{y=-\sqrt{25-x^2}}^{\sqrt{25-x^2}} \,dy\,dx \quad (5x-8y) \)
\(\large \int_{-5}^5 \,dx \quad [5xy-4y^2]_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}} \)
\(\large \int_{-5}^5 \,dx \quad [5x \sqrt{25-x^2}-4(25-x^2)] - [5x (-\sqrt{25-x^2})-4(25-x^2)] \\ \large = \int_{-5}^5 \,dx \quad 10x \sqrt{25-x^2}\)
\(\large = [(\frac{2}{3})(-\frac{1}{2}) 10x (25-x^2)^{3/2} ]_{-5}^5 \color{red}{= 0} \)

of course sith's polar switch is still superior and empty didn't even have to lift a pen to do this

the Green thing is really interesting. but firstly, i don't see how the field is conservative. if we re-cast the integral as \(\iint 6x-8y dA = \iint \nabla \times F \, dA \) then \( \nabla \times \mathbf{F} \ne 0\). is it not a case of the symmetry just creating a nil answer here?

furthermore, if one were clever enough to think Green in the first place, how would you reverse engineer the field from the curl to do the line integral? or am i being really dumb again?

Answer 7

I take back my comment about possibly using Green's theorem (though a tentative take-back). I just happened to wonder about how the double integral form somewhat resembled the one that equates it to a line integral. We would probably need some extra information about \(\bf F\) is we want to get anywhere.

Answer 8

Ok two things, first thing I will address is how to take this integral through the Green's theorem path (I just do it with Stokes' theorem since that's just how I think of it, same thing) And the other thing I will address later on is the conservative field comment I made. \[\iint 6x-8y dA = \iint \nabla \times F dA \]
So in order to make that substitution, I'm really saying that this is the curl of some vector field
\[6x-8y=\nabla \times F\]

More explicitly, let's just calculate the terms starting with \(\vec F = \langle F_x, F_y, 0 \rangle \)
\[\nabla \times F = \langle 0, 0, \frac{\partial F_y }{\partial x} - \frac{\partial F_x}{\partial y} \rangle \]

So now we equate this component with our curl (Yeah, cross products are awkward for 2D since you have to put in extra 3D garbage in with it, oh well, tensors are better :P)

\[ \frac{\partial F_y }{\partial x} - \frac{\partial F_x}{\partial y} = 6x-8y\] This next step is completely arbitrary to break it up into what's "obvious" although I'm pretty sure it doesn't matter how we break it up, not sure:

\[ \frac{\partial F_y }{\partial x} = 6x\]
\[ \frac{\partial F_x}{\partial y} = 8y\]

So finding one possible choice of F is not hard:

\[\vec F = \langle 4y^2, 3x^2, 0 \rangle \]

Now we are able to take this step further:
\[\iint 6x-8y dA = \iint \nabla \times F dA = \oint F \cdot dS= 0 \]

The path of our line integral is the boundary of a circle with radius 5, \(\vec c(t) = \langle 5 \cos t, 5 \sin t \rangle\) so we could proceed with this integral:

\[\int_0^{2 \pi} \vec F(x(t),y(t)) \cdot \vec c'(t) dt\]
\[\int_0^{2 \pi} 4*5^2 \sin^2(t) * (-5) \sin (t) + 3*5^2 \cos^2(t) *5 \cos(t) dt\]

which is an odd power of a trig function over an entire period so it should be zero, assuming everything I did was right, so this is at least how you could do it.

So let's move on to the second thing, the conservative field comment. I guess in order to be conservative, the line integral for all closed paths must be equal to 0, in other words the integral only depends on the state of the two end points and not the path it takes. So yeah I guess I goofed, it might not be. If I have any more mistakes above or you can clarify anything please do.