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Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

Write an equation for the polynomial graphed below

Anit [1.1K]

The expression for the polynomial graphed will be y(x) = (x + 3)(x - 1 )(x - 4 ).

<h3>How to factor the polynomial?</h3>

From the graph, the zeros of the polynomial of given graph are:

x = -3

x = 1

x = 4

Equate the above equations to zero

x + 3 = 0

x - 1 = 0

x - 4 = 0

Multiply the equations

(x + 3)(x - 1 )(x - 4 ) = 0

Express as a function gives;

y = (x + 3)(x - 1 )(x - 4 )

Hence, the factored form of the polynomial will be y = (x + 3)(x - 1 )(x - 4 ) .

Read more about polynomials at:

brainly.com/question/4142886

#SPJ1

6 0

1 year ago

Please help answer this :)

jolli1 [7]

Answer:

ok

Step-by-step explanation:

yeah

8 0

2 years ago

Select the equation that contains the points (-1, 8) and (3, 4)

bezimeni [28]

None of above.The answer is a line with this form:y - 4= - 1 (x - 3)

8 0

3 years ago

Given that the measure of ∠x is 90°, and the measure of ∠y is 53°, find the measure of ∠z.

jeka94

To find the answer we can use the angles sum of triangle which is 180° to find ∠z by deducting the other two angles from 180°:

180°-53°-90°
=37°

Therefore ∠z=37°.

Hope it helps!

6 0

3 years ago

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