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Over [174]
3 years ago
11

Solve the system by elimination. -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5

Mathematics
1 answer:
KengaRu [80]3 years ago
8 0
-2x + 2y +  3z = 0 → 2x - 2y -   3z = 0 → 2x - 2y - 3z = 0
-2x -   1y +  1z = -3 → 2x + 1y -    1z = 3 → 2x + 1y -  1z = 3
2x +   3y + 3z = 5 →  2x + 3y + 3z = 5            -3y - 2z = -3

-2x + 2y +  3z = 0 → 2x - 2y -   3z = 0
-2x -   1y +  1z = -3 → 2x + 1y -    1z = 3 → 2x + 1y -   1z = 3
2x +   3y + 3z = 5 →  2x + 3y + 3z = 5 → 2x + 3y + 3z = 5
                                                                           -2y - 4z = -2

-3y - 2z = -3 → -6y - 4z = -6
-2y - 4z = -2 → -2y - 4z = -2
                                 -4y = -4
                                  -4     -4
                                     y = 1

  -3y - 2z = -3
-3(1) - 2z = -3
   -3 - 2z = -3
 + 3          + 3
         -2z = 0
          -2    -2
            z = 0

    -2x + 2y + 3z = 0
-2x + 2(1) + 3(0) = 0
       -2x + 2 + 0 = 0
             -2x + 2 = 0
                    - 2  - 2
                    -2x = -2
                     -2     -2
                       x = 1
              (x, y, z) = (1, 1, 0)
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The two parabolas intersect for

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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

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\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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