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ivolga24 [154]
3 years ago
14

Determine the value of e.

Mathematics
1 answer:
Softa [21]3 years ago
7 0

9514 1404 393

Answer:

  A)  16

Step-by-step explanation:

The ratio of long-side to short-side is the same for all of these similar triangles, so we have ...

  e/8 = 8/4

  e = 8(8/4) = 8(2) = 16

The value of e is 16.

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Answer:

The integral of the volume is:

V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

The result is: V = 78.97731

Step-by-step explanation:

Given

Curve: x^2 + 4y^2 = 4

About line x = 2 --- Missing information

Required

Set up an integral for the volume

x^2 + 4y^2 = 4

Make x^2 the subject

x^2 = 4 - 4y^2

Square both sides

x = \sqrt{(4 - 4y^2)

Factor out 4

x = \sqrt{4(1 - y^2)

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x = \sqrt{4} * \sqrt{(1 - y^2)

x = \±2 * \sqrt{(1 - y^2)

x = \±2 \sqrt{(1 - y^2)

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x_1 = -2 \sqrt{(1 - y^2)}\ and\ x_2 = 2 \sqrt{(1 - y^2)}

Rotate about x = 2 implies that:

r = 2 - x

So:

r_1 = 2 - (-2 \sqrt{(1 - y^2)})

r_1 = 2 +2 \sqrt{(1 - y^2)}

r_2 = 2 - 2 \sqrt{(1 - y^2)}

Using washer method along the y-axis i.e. integral from 0 to 1.

We have:

V = 2\pi\int\limits^1_0 {(r_1^2 - r_2^2)} \, dy

Substitute values for r1 and r2

V = 2\pi\int\limits^1_0 {(( 2 +2 \sqrt{(1 - y^2)})^2 - ( 2 -2 \sqrt{(1 - y^2)})^2)} \, dy

Evaluate the squares

V = 2\pi\int\limits^1_0 {(4 +8 \sqrt{(1 - y^2)} + 4(1 - y^2)) - (4 -8 \sqrt{(1 - y^2)} + 4(1 - y^2))} \, dy

Remove brackets and collect like terms

V = 2\pi\int\limits^1_0 {4 - 4 + 8\sqrt{(1 - y^2)} +8 \sqrt{(1 - y^2)}+ 4(1 - y^2)  - 4(1 - y^2)} \, dy

V = 2\pi\int\limits^1_0 { 16\sqrt{(1 - y^2)} \, dy

Rewrite as:

V = 16* 2\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

Using the calculator:

\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy = \frac{\pi}{4}

So:

V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy

V = 32\pi * \frac{\pi}{4}

V =\frac{32\pi^2}{4}

V =8\pi^2

Take:

\pi = 3.142

V = 8* 3.142^2

V = 78.97731 --- approximated

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