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Please help me solve this problem,

$$\iint_S xy \;\mathrm dS$$

Where $S$ is the boundary of the region enclosed by the cylinder $x^2 + z^2 = 1$ and the planes $y = 0$ and $x + y = 2$.

I have tried this problem multiple times with no success. With that being said, I know the best way to solve this problem is to separate the integral into 3 separate boundaries (Top of the cylinder, bottom of the cylinder, and laterally). Thanks in advance

Graham Kemp
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b-rawld
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1 Answers1

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Indeed, it is a good idea to separate the integral on the three surfaces $S_1$, $S_2$, $S_3$, with $S=S_1 \cup S_2 \cup S_3$.

First, lets parametrize the surfaces:

  • $S_1$: $x=x,y=0,z=z$, with $(x,z)\in D:=\{(x,z)|x^2+z^2\le 1\}$, $\|r_x \times r_z\|=1$;
  • $S_2$: $x=\sin(t),y=y,z=\cos(t)$, with $0\le y \le 2-sin(t)$ and $0\le t\le 2\pi$, $\|r_t \times r_y\|=1$;
  • $S_3$: $x=x,y=2-x,z=z$, with $(x,z)\in D:=\{(x,z)|x^2+z^2\le 1\}$, $\|r_x \times r_z\|=\sqrt{2}$;

Now, we can start working the integral:

$$ \iint_{S_1 \cup S_2 \cup S_3}xy dS = \iint_D 0 dA +\int_0^{2\pi} \int_0^{2-\sin(t)}y \sin(t) dy dt +\iint_D x(2-x) \|r_x \times r_z\| dA $$

The first term equals $0$, the second $-2\pi$, and the third:

$$ \iint_D x(2-x) \|r_x \times r_z\| dA = \int_{0}^{2\pi}\int_0^1\sqrt{2}r\sin(t)(2-r\sin(t))r dr dt = \frac{-\pi \sqrt{2}}{4}. $$

So in the end: $$ \iint_{S=S_1 \cup S_2 \cup S_3}xy dS = -2\pi - \frac{\pi \sqrt{2}}{4}. $$

Kuifje
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