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grigory [225]
3 years ago
7

Can someone just check this to make sure it is correct and please be honest because my grade depends on this

Mathematics
1 answer:
Anarel [89]3 years ago
5 0

Answer:

Your answers are right

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Rewrite the following integral in spherical coordinates.​
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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}

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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

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