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3 years ago

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

1 answer:

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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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 $2(x^2-4)$ .

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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:

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

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

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