0.7 repeating as a fraction

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{{81 -y^2} }{2} \end {bmatrix} \ dxdy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

4 0

3 years ago

The segment show below form a triangle<br> True or false?

seropon [69]

Answer:

true

Step-by-step explanation:

because it would be

5 0

3 years ago

Help with both please ;-;

Schach [20]

Volume = l*b*h

1. Ans 24 unit cubes

Because it had 6 unit cubes as length, 2 unit cunes as breadth and 2 unit cubes as height.

2. Volume = l*b*h
= 30*15*4
=1800 cubic feet

4 0

3 years ago

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A sector with a central angle measure of 200 degrees has a radius of 9 cm. What is the area of the sector?

Sergio [31]

Answer:

$\boxed{Area\ of\ sector = 141.4\ cm^2}$

Step-by-step explanation:

Radius = r = 9 cm

Angle = θ = 200° = 3.5 radians

Now,

$Area \ of \ sector = \frac{1}{2} r^2 \theta$

Area = 1/2 (9)²(3.5)

Area = 1/2 (81)(3.5)

Area = 282.7 / 2

Area of sector = 141.4 cm²

6 0

3 years ago

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Solve the following equations. 2x - 7 = 11<br>help me please it's urgent

elena-s [515]

Answer:

X=9

Step-by-step explanation:

2x-7=11

Add 7 each side

2x - 7 +7 =11 + 7

2x=18

Divide by 2 each side

2x/2=18/2

x=9

4 0

2 years ago

Read 2 more answers