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drek231 [11]
2 years ago
6

Calculus Problem

Mathematics
1 answer:
Roman55 [17]2 years ago
7 0

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}}

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Answer:

1i) nth term = 12(-b/4)^(n-1)

ii) nth term = 3(-1/9)^(n-1)

2i) Sixth term= (-3b^5)/256

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Question:

The complete question as found on brainly (question-10746111):

Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.

i) 12,-3b,......sixth term

ii) 3,-1/3,..........;8th term?

Step-by-step explanation:

1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.

i) 12,-3b,......;sixth term

1st term = a= 12

r = common ratio = 2nd term /1st term

r = -3b/12 = -b/4

Then we would find the nth term

See attachment for details

ii) 3,-1/3,..........;8th term

1st term = a= 3

r = common ratio = 2nd term /1st term

r = (-⅓)/3 = (-⅓)(⅓) = -1/9

Then we would find the nth term

See attachment for details

2) we are to determine the sixth term in (i) and eight term in (ii)

See attachment for details

Sixth term= (-3b^5)/256

Eight term = -1/1594323

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