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Evaluate $$I=\iint\limits_R \sin \left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\, dA,$$ where $R$ is the triangle with vertices $(0,0),(2,0)$ and $(1,1)$.

Hint: use $u=\dfrac{x+y}{2},v=\dfrac{x-y}{2}$.

Can anyone help me with this question I am very lost. Please help I know you can make the intergal $\sin(u)\cos(v)$, but then what to do?

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    $\begingroup$ Hint: Use the hint. What happens to the triangular integration region then? $\endgroup$ – Lord Soth Jun 29 '13 at 20:13
  • $\begingroup$ it would become sin(u)cos(v) this is where i get lost. $\endgroup$ – jain smit Jun 29 '13 at 20:15
  • $\begingroup$ Do you know what a change of coordinates is and how you can use them to simplify integrals? Does the word Jacobian mean anything to you? $\endgroup$ – James Jun 29 '13 at 20:16
  • $\begingroup$ No i do not get what you are saying. Hint? $\endgroup$ – jain smit Jun 29 '13 at 20:17
  • $\begingroup$ Did you learn Jacobian in your class? or you missed it? $\endgroup$ – Shuhao Cao Jun 29 '13 at 20:42
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There is another way to do this. You might notice that the integrand is

$$2 \sin{\left(\frac{x+y}{2}\right)} \cos{\left(\frac{x-y}{2}\right)} = \sin{x} + \sin{y}$$

You may then integrate this over the triangle directly:

$$\frac12 \int_0^1 dx \, \int_0^x dy \, [\sin{x} + \sin{y}] + \frac12 \int_1^2 dx \, \int_0^{2-x} dy \, [\sin{x} + \sin{y}] $$

Note that I formed the integration boundaries from the equations of the lines formed from the vertices of the triangle. Note also that I had to break this in two: one for the left side and one for the right.

You may then evaluate this in terms of single integrals by integrating over $y$; I get

$$\frac12\int_0^1 dx \, [x \, \sin{x} + 1 - \cos{x}] + \frac12\int_1^2 dx \, [(2-x) \sin{x} + 1-\cos{(2-x)}]$$

I will let you take it from here.

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  • $\begingroup$ I think you missed a $\frac{1}{2}$ factor in your first equation and I didn't understand where did the $x\sin x$ appear from? $\endgroup$ – Paracosmiste Jun 29 '13 at 21:05
  • $\begingroup$ @metacompactness: the factor of $1/2$ should show up in an edit - it was a mistake. The $x \sin{x]$ comes form integrating $\sin{y}$ over $x$. $\endgroup$ – Ron Gordon Jun 29 '13 at 21:06
  • $\begingroup$ Actually, it's integrating $\sin x$ over $y$ but I got your point. $\endgroup$ – Paracosmiste Jun 29 '13 at 21:10
  • $\begingroup$ Where did you get the integral 2sin()cos()? $\endgroup$ – jain smit Jun 29 '13 at 21:21
  • $\begingroup$ @jainsmit: not sure what you mean $\endgroup$ – Ron Gordon Jun 29 '13 at 21:47
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Here is how you advance

1) Solve the system

$$u=\dfrac{x+y}{2},\,v=\dfrac{x-y}{2} $$

for $x$ and $y$.

2) Find the Jacobian

$$ J=\begin{bmatrix} \dfrac{\partial x}{\partial u} \quad \dfrac{\partial x}{\partial v} \\ & \\ \dfrac{\partial y}{\partial u} \quad \dfrac{\partial y}{\partial v} \end{bmatrix}. $$

3) $$ dxdy = |J|dudv .$$

4) Find the limits of integration for $u$ and $v$.

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  • $\begingroup$ so would you just take the integral with J alone? $\endgroup$ – jain smit Jun 29 '13 at 23:13
  • $\begingroup$ @BabakS. Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Jul 2 '13 at 15:27
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Following the hint we obtain a new coordinate system rotated $\frac{\pi}{4}$ clockwise. See the following picture.

enter image description here

A unit area in the new system is the double of the old one. Thus the integral changes to: $$2 \int_0^{1} \int_v^1{\sin u \cos v \, \mathrm {d}u \, \mathrm {d}v}$$ which you can easily finish.

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$\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=\frac{1}{2}\left(\sin x+\sin y\right)$

The line joining $(0,0)$ and $(2,0)$ has an equation $y=0$ and $0\leq x\leq 2$

The second line: $y=-x+2$

The third line: $y=x$

The integral becomes: $$I=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}+\int\limits_{1}^{2}\int\limits_{0}^{2-x}\right)(\sin x+\sin y)dy~dx=\frac{1}{2}\left(\int\limits_{0}^{1}\int\limits_{0}^{x}(\sin x+\sin y)dy~dx+\int\limits_{1}^{2}\int\limits_{0}^{2-x}(\sin x+\sin y)dy~dx\right)=\frac{1}{2}\left(\int\limits_0^1\left[y\sin x-\cos y\right]_0^xdx+\int\limits_1^2\left[y\sin x-\cos y\right]_0^{2-x}dx\right)=\frac{1}{2}\left(\left[\sin x-x\cos x+x-\cos x\right]_0^1+\left[(x-2)\cos x-\sin x+x+\sin(2-x)\right]_1^2\right)$$

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  • $\begingroup$ why did you make sin and cos into sin+sin? $\endgroup$ – jain smit Jun 29 '13 at 21:16
  • $\begingroup$ @jainsmit It's called linearization and it is often used in integration. $\endgroup$ – Paracosmiste Jun 29 '13 at 21:23
  • $\begingroup$ ok I got my answer as .545351 is that right? ans also can you finish the integral so i can compare my work to yours? $\endgroup$ – jain smit Jun 29 '13 at 22:47
  • $\begingroup$ I finished the double integral but I don't have a calculator right now to find the approximate value. You can always use the Wolfram integrator to find the integrals. $\endgroup$ – Paracosmiste Jun 30 '13 at 9:57

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