2
$\begingroup$

Use Green's Theorem to find the work done by the force $\mathbf{F}(x,y)=x(x+y)\mathbf{i}+xy^2\mathbf{j}$ in moving a particle from the origin along the $x$-axis to $(1,0)$, then along the line segment to $(0,1)$, and back to the origin along the $y$-axis.

So I was able to find $\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y}$ to be $y^2 -x$ and I integrated that with respect to $y$ and $x$ by using $y= 1-x$ as my upper bound and $y=0$ as my lower bound, and $0 < x < 1$ for my $x$ integral. but it came out to $-\frac{7}{36}$, and the answer is $-\frac1{12}$. I'm not sure if I'm doing something fundamentally wrong here or if its a calculation error. I checked it twice. How do I do it correctly?

$\endgroup$
3
$\begingroup$

Yes, you are correct, by Green's Theorem, you should evaluate $$\int_{x=0}^1\int_{y=0}^{1-x}(y^2-x)dydx=\int_{x=0}^1\left[\frac{y^3}{3}-xy\right]_{y=0}^{1-x}dx=\int_{x=0}^1\left(\frac{(1-x)^3}{3}-x(1-x)\right)dx\\ =\frac{1}{3}\int_{t=0}^1t^3 dt-\int_{x=0}^1 x(1-x)dx.$$ Can you take it from here?

$\endgroup$
  • $\begingroup$ well apparently not, I tried and got it wrong. but at least I got the method right. I did end up with that integral on the right side though. $\endgroup$ – 2316354654 Dec 5 '17 at 7:44
  • 1
    $\begingroup$ In the first integral, you may let $t=1-x$. $\endgroup$ – Robert Z Dec 5 '17 at 7:48
  • $\begingroup$ what I did was expand everything before integrating and ended up wtih 3 variables. It was probably a simple arithmetic error $\endgroup$ – 2316354654 Dec 5 '17 at 7:50
  • 1
    $\begingroup$ Yes, expanding the cube could be the source of your error. $\endgroup$ – Robert Z Dec 5 '17 at 7:56
  • $\begingroup$ i just found my arithmetic error. now i got -1/12. you were right, it was the cube expansion. $\endgroup$ – 2316354654 Dec 5 '17 at 7:59
3
$\begingroup$

\begin{align}\int_0^1 \int_0^{1-x} (y^2-x) \, dy\, dx &= \int_0^1 \left[ \frac{y^3}{3}-xy \right]_{y=0}^{y=1-x}\, dx \\ &=\int_0^1 \frac{(1-x)^3}{3}-x(1-x) \, dx \\ &= \int_0^1 \frac{(1-x)^3}{3} - x+x^2 \, dx \\ &= \left[ -\frac{(1-x)^4}{12}-\frac{x^2}{2}+\frac{x^3}{3}\right]_{x=0}^{x=1} \\ &= \left[ -\frac12+\frac13+\frac1{12} \right]\\ &=\left[\frac{-6+4+1}{12} \right]\\ &=-\frac1{12}\end{align}

Your method seems correct but most likely you make a careless mistake somewhere.

$\endgroup$
2
$\begingroup$

$$\int_{y=0}^1\int_{x=0}^{1-y}(y^2-x)dxdy = \int_{x=0}^1(y^2x-\frac{x^2}{2})|_{0}^{(1-y)}dy$$

If you expand you get $$=\frac{1}{2}\int_{y=0}^1(-2y^3+y^2+2y-1)dy$$

$$=\frac{1}{2}\left(\frac{-2y^4}{4}+\frac{y^3}{3}+\frac{2y^2}{2}-y\right)|_{0}^{1}$$

$$=-\frac{1}{12}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.