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2 years ago

Given: m EH =85°, m∠EYV=35°. Find: m EV .

Mathematics

1 answer:

avanturin [10]2 years ago

5 0

Answer:

155°

Step-by-step explanation:

∠EYV is half the difference of arcs EV and EH

(1/2)(EV -EH) = ∠EYV

(1/2)(EV -85°) = 35° . . . . fill in the given values

EV -85° = 70° . . . . . . . . .multiply by 2

EV = 155° . . . . . . . . . . . . add 85°

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Step-by-step explanation:

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The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

$\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx$

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$v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx$

$v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx$

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$v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx$

Evaluate the exponents

$v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx$

Simplify like terms

$v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx$

Factor out 8

$v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx$

Integrate

$v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0$

$v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0$

Substitute 2 and 0 for x, respectively

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])$

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])$

$v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]$

$v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]$

Take LCM

$v = 8\pi [ \frac{16*4- 16 * 3}{12}]$

$v = 8\pi [ \frac{64- 48}{12}]$

$v = 8\pi * \frac{16}{12}$

Simplify

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$v = \frac{32\pi}{3}$

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$v=\frac{32*22}{3*7}$

$v=\frac{704}{21}$

$v=33.52$

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