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maria [59]
3 years ago
7

Find the mass of the wire formed by the intersection of the sphere x 2 + y 2 + z 2 = 1 and the plane x + z = 0 if the density of

the wire is 4x 2 grams per unit length.
Mathematics
1 answer:
Marat540 [252]3 years ago
6 0
<span>Answer: x²+(â’xâ’z)²+z² = 1 → 2x²+2zx+(2z²â’1) = 0 Solve as a quadratic : x = { â’2z ± âš(4z²â’8(2z²â’1) } / 4 = { â’z ± âš(2â’3z²) } / 2 A more manageable form can be derived by setting z = âš(2/3).sint to remove âš This gives x = â’âš(1/6).sint + âš(1/2).cost (can omit ± sign since â’ is covered by Ď€â’t) Using x+y+z = 0 gives y = â’xâ’z = â’âš(1/6)sint â’ âš(1/2).cost So a parameterisation in terms of t â [0,2Ď€] is x = â’âš(1/6).sint + âš(1/2).cost, y = â’âš(1/6).sint â’ âš(1/2).cost, z = âš(2/3).sint (ds/dt)² = (â’âš(1/6).costâ’âš(1/2).sint)² + (â’âš(1/6).cost+âš(1/2).sint)² + (âš(2/3).cost)² = 1 â´ â« Ďds = â« ( â’âš(1/6).sint + âš(1/2).cost )²dt, [t=0,2Ď€] = 2Ď€/3</span>
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Alex777 [14]

Answer:

<em>The area of the shaded region is 3.</em>

Step-by-step explanation:

Since point E lies halfway between AB and BC, the area of the shaded region (As) consists of two identical triangles with base equal to AB and height equal to half the measure of BC:

A_s=2 * A_t

Where At is the area of each triangle.

\displaystyle A_t=\frac{AB*BC/2}{2}

\displaystyle A_t=\frac{AB*BC}{4}

We know AB=3 and BC=2, thus:

\displaystyle A_t=\frac{3*2}{4}=\frac{6}{4}

Simplifying:

\displaystyle A_t=\frac{3}{2}

Finally:

\displaystyle A_s=2 * \frac{3}{2}

A_s=3

The area of the shaded region is 3.

Note the area of the shaded region is half the area of the rectangle.

3 0
2 years ago
A cone and its dimensions are shown in the diagram. Which measurement is closest to the volume of the cone in cubic inches?
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(The ^2 means squared, or to the second power)

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Answer:

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