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Ipatiy [6.2K]
3 years ago
8

Convert the integral below to polar coordinates and evaluate the integral.

Mathematics
1 answer:
Lelechka [254]3 years ago
6 0

Answer:

\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}}  xy \, dxdy =  \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4}    r^3 cos(\theta)\sin(\theta) \,\,  drd\theta = 16

Step-by-step explanation:

We are trying to evaluate this integral.

\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}}  xy \,\,dxdy

The first thing that we have to do is understand this region in the plane.

\{ (x,y) \in \mathbb{R} :  0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2}   \}

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4}  \,\,\, 0\leq r \leq 4  \}

Now remember that when we do the polar transformation we use the following formula

\int\limits_{a}^{b} \, \int\limits_{c}^{d}    f(x,y) \,dxdy =  \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2}    r* f(rcos(\theta),rsin(\theta))  \,drd\theta

Then our integral would be

\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}}  xy \, dxdy =  \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4}    r^3 cos(\theta)\sin(\theta) \,\,  drd\theta = 16

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