Question

Evaluate the integral directly:
I = int((xy + x) dx + (x2 + xy) dy) on C
where C is the boundary of the region lying between the curves y = x and y =sqrt(x)
0 <= x <= 1.

Answer 1

I think we can apply Green's theorem here.

Answer 2

I will write down what I have (You need to look at the theorem):
\[I=\iint_R ({\partial \over \partial x}(x^2+xy)-{\partial \over \partial y}(xy+x))dA=\iint_R(2x+y-x)dA\]
\[=\iint_{R}(x+y)dA\].
The region R is defined by \(0\le x \le 1\) and \(x\le y\le \sqrt{x}\). Hence
\[I=\int\limits_0^1 \int\limits_{x}^{\sqrt{x}}(x+y)dy dx.\]

Answer 3

This integral is easy to compute. I got \(I=\frac{3}{20}\).

Answer 4

Thanks AnwarA, that's helpful, but I also have to calculate the integral withough green's theorem and I'm not sure how to do that...

Answer 5

you can compute 2 simple integral

one along x=t, y=t t:0->1... this gives 3/2


one along x=t and t=sqrt(t) t:1->0 gives -27/20

3/2-27/20=3/20

Answer 6

\[y=\sqrt{t}\]

Answer 7

one along \[x=t\text{ and }y=\sqrt{t} \text{ from }t:1\to 0 \text{ gives }\frac{-27}{20}\]

Answer 8

the first one is
x=t ,y=t so dx=dt, dy=dt

\[\int\limits_{0}^{1}((t\cdot t + t) dt + (t^2 + t\cdot t) dt) \]

\[\int\limits_{0}^{1}(3t^2+ t) dt=\frac{3}{2} \]