The region bounded by the given curve is rotated about the specified axis. find the volume v of the

10 answers
Question:

The region bounded by the given curve is rotated about the specified axis. find the volume v of the resulting solid by any method. x^2 + (y − 2)^2 = 4; about the y-axis

Answers

X^2+(y-2)^2=4 is a circle centered at (0,2) with radius 2.
Thus, when you rotate it around the y-axis (around x=0),
you get a sphere with radius 2.

((which is centered at (x,y,z)=(0,2,0), if you take y to be
the vertical axis ... not that you need to know this for volume))


Volume of a sphere formula: V = (4/3)πr^3.

[tex]x^2+(y-2)^2=4\implies y=2\pm\sqrt{4-x^2}[/tex]

describes a circle of radius 2 centered at (0,2). A revolution of this region about the y-axis would give you a sphere of radius 2, so the volume would be [tex]\dfrac43\pi\times2^3=\dfrac{32}3\pi[/tex].

[tex]\bf x^2+(y-4)^2=16\implies x^2=16-(y-4)^2
\\\\\\
x=\sqrt{16-(y-4)^2}\implies x=\sqrt{16-(y^2-8y+16)}
\\\\\\
x=\sqrt{8y-y^2}\\\\
-------------------------------\\\\
0=\sqrt{8y-y^2}\implies 0=8y-y^2\implies 0=y(8-y)
\implies 
y=
\begin{cases}
0\\
8
\end{cases}\\\\
-------------------------------\\\\
\displaystyle \stackrel{\textit{disk method}}{\int\limits_{0}^8~\pi (\sqrt{8y-y^2})^2\cdot dy}\implies \pi \int\limits_{0}^8~ 8y-y^2\cdot dy[/tex]

[tex]\bf \displaystyle \pi \int\limits_{0}^8~8y\cdot dy - \pi \int\limits_{0}^8~y^2\cdot dy\implies \pi \left[ 4y^2~-~\cfrac{y^3}{3} \right]_{0}^8
\\\\\\
\pi \left[ \left[ 256-\cfrac{51}{3} \right]~-~[0] \right]\implies \pi \cdot \cfrac{256}{3}\implies \cfrac{256\pi }{3}\implies 85\frac{\pi }{3}[/tex]

0

Step-by-step explanation:

y = −x2 + 10x − 24,  

y = 0;  

0= -x2 + 10x - 24 /0

0 = 0

0

Step-by-step explanation:

y = −x2 + 10x − 24,

y = 0;

0= -x2 + 10x - 24 /0

0 = 0

[tex]\frac{128\pi}{3}[/tex] cubic units

Step-by-step explanation:

We are given that a region bounded by a curves [tex]x=(y-7)^2[/tex]

x=4 and about y=5

We have to find the value of volume of the resulting solid

To find the volume of resulting solid we are using cylinder method

Substitute the values of x then we get

[tex](y-7)^2=4[/tex]

[tex]y-7=\pm2[/tex]

y-7=2  and y-7=-2

y=7+2=9 and y=-2+7=5

Radius of cylinder =r=y-5

and height =h=[tex]4-(y-7)^2=4-y^2-49+14 y=-y^2+14 y -45[/tex]

Using cylinder method and integrate along y -axis from y=5 to y=9

Volume =[tex]\int_{a}^{b}2\pi r h dy=\int_{5}^{9}2\pi(y-5)(-y^2+14 y -45) dy[/tex]

volume=[tex]2\pi\int_{5}^{9}(-y^3+19y^2-115y+225)dy[/tex]

Volume =[tex]2\pi[-\frac{y^4}{4}+19\frac{y^3}{3}-115\frac{y^2}{2}+225y]^9_5[/tex]

Volume =[tex]2\pi[-\frac{6561}{4}+\frac{625}{4}+19(\frac{729-125}{3})-115(\frac{81-25}{2})+225(9-5)][/tex]

Volume=[tex]2\pi(-1484+\frac{11476}{3}-3220+900)[/tex]

Volume =[tex]2\pi(\frac{11476}{3}-3804)[/tex]

Volume =[tex]2\pi(\frac{11476-11412}{3})=\frac{128\pi}{3}[/tex]

Hence, volume of resulting solid =[tex]\frac{128\pi}{3}[/tex]cubic units

[tex]V=\frac{512\pi }{15}[/tex]

Step-by-step explanation:

First thing I did was graph this on my calculator to find the upper and lower bounds.  They were x = 5, 9.  Then I applied the disk method of integration.  From this I know know I use y = x equations, x intervals, I need a value for R, and then I need a value for r.  R represents the height of the representative rectangle which is perpendicular to the axis of rotation, so the height of that rectangle starts at y = 0 and meets the curve.  So

[tex]R=-x^2+14x-45[/tex]

Since there is no "hole" or space between the graph and the axis of rotation, r=0.  Putting that into the volume formula:

[tex]V=\pi \int\limits^9_5 {[-x^2+14x-45]^2-[0]^2} \, dx[/tex]

To simplify I have to multiply that polynomial by itself.  Doing that gives us:

[tex]V=\pi \int\limits^9_5 {x^4-28x^3+286^2-1260x+2025} \, dx[/tex]

Integrating gives us:

[tex]V=\pi [\frac{x^5}{5}-\frac{28x^4}{4}+\frac{286x^3}{3}-\frac{1260x^2}{2}+2025x[/tex] which we integrate from 5 to 9

Doing a bit of simplification on that:

[tex]V=\pi [\frac{x^5}{5}-7x^4+\frac{286x^3}{3}-630x^2+2025x][/tex] from 5 to 9

Applying the First Fundamental Theorem of Calculus we get that when we sub in a 9 for x then a 5 for x:

[tex]V=\pi(\frac{12879}{5}-\frac{7625}{3})[/tex]

which subtracts to

[tex]V=\frac{512\pi }{15}[/tex]

Hence the volume of the solid is: [tex]\dfrac{256}{3}\pi[/tex]

Step-by-step explanation:

We will use the washer method to compute the volume of the solid formed after rotating the given figure around y-axis as:

Clearly the resulting solid is a sphere.

Now the washer method is given as:

Volume= [tex]\int\limits^8_0 {\pi x^2} \, dy[/tex]

Now we are given the equation as:

[tex]x^2 + (y-4)^2 = 16[/tex]

so we get the value of [tex]x^2[/tex] as:

[tex]x^2=16-(y-4)^2[/tex]

Now the volume is computed as:

Volume=  [tex]\int\limits^8_0 {\pi (16-(y-4)^2 ) } \, dy[/tex]

           =  [tex]\pi \int\limits^8_0 {16-(y-4)^2} \, dy\\\\=\pi [16y-\dfrac{(y-4)^3}{3}][/tex] from 0 to 8

Which is given by putting the limits.

Volume=  [tex]\pi [16\times 8-\dfrac{(8-4)^3}{3}-(16\times 0-\dfrac{(0-4)^3}{3})]\\\\=\pi [128-\dfrac{4^3}{3}+\dfrac{(-4)^3}{3}]\\\\=\pi[128-2\times \dfrac{4^3}{3}]\\\\=\pi[128-\dfrac{128}{3}]\\\\=\pi\times 128\times [1-\dfrac{1}{3}]\\\\=\pi\times 128\times \dfrac{3-1}{2}\\\\\pi\times 128\times \dfrac{2}{3}\\\\=\dfrac{256}{3}\pi[/tex]

Hence the volume of the solid is: [tex]\dfrac{256}{3}\pi[/tex]

V  = 23π/6

Step-by-step explanation:

V = 2π ∫ [a to b] (r * h) dx

y = −x² + 23x − 132

y = −(x² − 23x + 132)

y = −(x − 11) (x − 12)

Parabola intersects x-axis (line y = 0) at x = 11 and x = 12 > a = 11, b = 12

r = x

h = −x² + 23x − 132

V = 2π ∫ [11 to 12] x (−x² + 23x − 132) dx

V  = 23π/6

Consider, pls, the suggested option (see the attachment).
[tex]The region bounded by the given curves is rotated about the specified axis. find the volume v of the[/tex]
[tex]The region bounded by the given curves is rotated about the specified axis. find the volume v of the[/tex]

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