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wolverine [178]
3 years ago
15

Find the area of the region that lies inside the first curve and outside the second curve. r = 6 − 6 sin θ, r = 6

Mathematics
1 answer:
yan [13]3 years ago
8 0
Each curve completes one loop over the interval 0\le t\le2\pi. Find the intersections of the curves within this interval.

6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi

The region of interest has an area given by the double integral

\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta

equivalent to the single integral

\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta

which evaluates to 9\pi+72.

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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 ratio of girls to boys in mrs tuners class is 3:4. there 9 girls in mrs tuners class. how many boys are in mrs tuners class
Aleonysh [2.5K]

Answer:

12

Step-by-step explanation:

You set up cross multiplication 3/4 x  9/x, you go 4x9=36 divided by 3 equals 12

7 0
2 years ago
Which function is graphed below?
lidiya [134]

Answer:

B: y= 3(1/3)^x

Step-by-step explanation:

I graphed it on Desmos

6 0
3 years ago
What are the solution(s) of the quadratic equation 98 – x2 = 0
DanielleElmas [232]
To solve first move the 98 to the other side by subtracting it from both sides. Then divide by - 1 on both sides to that the x^2 is no longer a negative. You are left with:

x^{2} = 98

Now square root both sides to get

x = +/-  \sqrt{98}

The two factors that make up 98 are 49 and 2. 49 can be square rooted. So we are left with these answers:

x = - 7 \sqrt{2} and x = 7 \sqrt{2}
4 0
3 years ago
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Pachacha [2.7K]
M would have a new coordinate of (4,5)
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3 years ago
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