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sp2606 [1]
3 years ago
9

Formula of arc length

Mathematics
2 answers:
vitfil [10]3 years ago
8 0
The formula is

(theta)/360 · 2(pi)r

theta is the given angle

r is radius of the circle
Dvinal [7]3 years ago
7 0
A formula of an arc length, starting from degrees is θ/360 * 2πr.

Hope this helps!

If you have any questions, please comment below.
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Louis bought packages of donuts. There were 4 donuts in each packages, and Louis gave 6 to his friend. Write an expression to sh
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4x-6

Step-by-step explanation:

4x-6

4 is for number of doughnuts in each package, 6 is for the taken doughnuts

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C (third graph, the parabola); To find if it is a function or not draw a vertical line on the graph. If the vertical line could never intercept the figure more than once anywhere on the graph then it is a function.

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5 0
3 years ago
Rewrite the following integral in spherical coordinates.​
lora16 [44]

In cylindrical coordinates, we have r^2=x^2+y^2, so that

z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}

correspond to the upper and lower halves of a sphere with radius \sqrt2. In spherical coordinates, this sphere is \rho=\sqrt2.

1 \le r \le \sqrt2 means our region is between two cylinders with radius 1 and \sqrt2. In spherical coordinates, the inner cylinder has equation

x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)

This cylinder meets the sphere when

x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)

which occurs at

\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi

where n\in\Bbb Z. Then \frac\pi4\le\phi\le\frac{3\pi}4.

The volume element transforms to

dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi

Putting everything together, we have

\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3

4 0
2 years ago
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