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Nikolay [14]
3 years ago
11

Find the volume of the solid bounded by z = 2 - x2 - y2 and z = 1. Express your answer as a decimal rounded to the hundredths pl

ace.
Mathematics
1 answer:
butalik [34]3 years ago
8 0

Answer:

The answer is "(\frac{\pi}{2})".

Step-by-step explanation:

z = 2 - x^2 - y^2.........(1) \\\\z = 1.............(2)  

Let add equation 1 and 2:

Using formula: x^2+y^2=1 \\\\

convert to polar coordinates

r=2

\theta \varepsilon (0=\pi)=z\\\\V=\int^{2\pi}_{\theta=0}\int^{1}_{\pi=0}\int^{z_{2}}_{z_1} r \ dr \d \theta\\\\

    =\int^{2\pi}_{0}\int^{1}_{0}  (Z_2-z_1)r  \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r)  \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0}  \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4})  \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\

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3 years ago
-2x + y =0 <br> 5x + 3y = -11
Ivahew [28]

Answer:

(-1, -2)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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<u>Algebra I</u>

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Step-by-step explanation:

<u>Step 1: Define Systems</u>

-2x + y = 0

5x + 3y = -11

<u>Step 2: Rewrite Systems</u>

-2x + y = 0

  1. Add 2x on both sides:                    y = 2x

<u>Step 3: Redefine Systems</u>

y = 2x

5x + 3y = -11

<u>Step 4: Solve for </u><em><u>x</u></em>

<em>Substitution</em>

  1. Substitute in <em>y</em>:                         5x + 3(2x) = -11
  2. Multiply:                                    5x + 6x = -11
  3. Combine like terms:                11x = -11
  4. Isolate <em>x</em>:                                   x = -1

<u>Step 5: Solve for </u><em><u>y</u></em>

  1. Define equation:                    y = 2x
  2. Substitute in <em>x</em>:                       y = 2(-1)
  3. Multiply:                                  y = -2

<u>Step 6: Graph Systems</u>

<em>Check the solution set.</em>

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