You want to find the volume inside the hemisphere $x^2+y^2+z^2=4$ (i.e. inside the sphere but above the plane $z=0$ ) and outside the cylinder $x^2+y^2=1$ . Call this region $R$ .

In cylindrical coordinates, we have

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_1^2\int_0^{\sqrt{4-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$

$\displaystyle=2\pi\int_1^2 r\sqrt{4-r^2}\,\mathrm dr$

$\displaystyle=-\pi\int_3^0\sqrt u\,\mathrm du$

(where $u=4-r^2$ )

$\displaystyle=\pi\int_0^3\sqrt u\,\mathrm du$

$=\dfrac{2\pi}3u^{3/2}\bigg|_0^3=2\sqrt3\,\pi$

7 0

3 years ago

PLEASE HELP ME! IS THIS CORRECT?

Amanda [17]

Answer:

yes

Step-by-step explanation:

3 0

3 years ago

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Find the solution to the system of equations by using either graphing or substitution.

Minchanka [31]

Answer: The answer is (-7,12)

Step-by-step explanation:

5 0

3 years ago

The length of XY the minor arc is 48 cm what is the circumference of Z

Amanda [17]

If found the image that accompanies this problem, the central angle was 60°. Since the circumference of a circle is always 360°, the minor arc represents 60°/360° of the circle.

48 cm / (60°/360°) = 48 cm / (1/6) = 48 cm * 6/1 = 48 cm * 6 = 288 cm

The circumference of circle Z is 288 cm.

4 0

3 years ago

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