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SpyIntel [72]
3 years ago
6

Use cylindrical coordinates to evaluate the triple integral ∭ where E is the solid bounded by the circular paraboloid z = 9 - 16

(x^2 + y^2) and the xy -plane.
Mathematics
1 answer:
4vir4ik [10]3 years ago
4 0

Answer:

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

Step-by-step explanation:

The Cylindrical coordinates are:

x = rcosθ, y = rsinθ and z = z

From the question, on the xy-plane;

9 -16 (x^2 + y^2) = 0 \\ \\  16 (x^2 + y^2)  = 9 \\ \\  x^2+y^2 = \dfrac{9}{16}

x^2+y^2 = (\dfrac{3}{4})^2

where:

0 ≤ r ≤ \dfrac{3}{4} and 0 ≤ θ ≤ 2π

∴

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} \int ^{9-16r^2}_{0} \ r \times rdzdrd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 z|^{z= 9-16r^2}_{z=0}  \ \ \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0} r^2 ( 9-16r^2})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0} \int ^{\dfrac{3}{4}}_{0}  ( 9r^2-16r^4})  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( \dfrac{9r^3}{3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3r^3}-\dfrac{16r^5}{5}})|^{\dfrac{3}{4}}_{0}  \ drd \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV = \int^{2 \pi}_{0}   ( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) d \theta

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}}) \theta |^{2 \pi}_{0}

\iiint_E  E \sqrt{x^2+y^2} \ dV =( 3(\dfrac{3}{4})^3}-\dfrac{16(\dfrac{3}{4})^5}{5}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{64}}-\dfrac{243}{320}})2 \pi

\iiint_E  E \sqrt{x^2+y^2} \ dV =(\dfrac{81}{160}})2 \pi

\mathbf{\iiint_E  E \sqrt{x^2+y^2} \ dV =\dfrac{81 \  \pi}{80}}

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Silicon wafers are scored and then broken into the many small microchips that will he mounted into circuits. Two breaking method
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Answer:

a

The estimate is  - 0.0265\le  K \le  0.0465

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Method B this is because the faulty breaks are less

Step-by-step explanation:

The number of microchips broken in method A is  n_1 = 400

The number of faulty breaks of method A is  X_1 = 32

 The number of microchips broken in method B is  n_2  = 400

 The number of faulty breaks of method A is  X_2 = 32

  The proportion of the faulty breaks to the total breaks in method A is

       p_1 = \frac{32}{400}

      p_1 = 0.08

 The proportion of the faulty to the total breaks in method B is

      p_2 =  \frac{28}{400}

     p_2 =  0.07

For this estimation the standard error is  

      SE =  \sqrt{ \frac{p_1 (1 - p_1)}{n_1}  + \frac{p_2 (1- p_2 )}{n_2} } }

  substituting values

       SE =  \sqrt{ \frac{0.08 (1 - 0.08)}{400}  + \frac{0.07 (1- 0.07 )}{400} } }

      SE = 0.0186

The z-values of confidence coefficient of 0.95 from the z-table is  

       z_{0.95} =  1.96

The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as

        K = [p_1 - p_2 ] \pm z_{0.95} * SE

substituting values

        K = [0.08 - 0.07 ] \pm 1.96 *0.0186

         K  =  - 0.0265 \ or  \ K  =  0.0465

The interval of the difference between proportions of improperly broken microchips for the two breaking methods is  

      - 0.0265\le  K \le  0.0465

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