What is the answer to 3/4 - 1/6

Mathematics

1 answer:

8_murik_8 [283]3 years ago

4 0

3/4-1/6
make bottom same
12

3/4 times 3/3=9/12
1/6 times 2/2=2/12

3/4-1/6=9/12-2/12=(9-2)/12=7/12

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

Lelechka [254]

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

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