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The region bounded by the given curve is rotated about the specified axis. Find the volume V of the resulting solid by any method.
x2 + (y − 4)2 = 16; about the y-axis

The region bounded by the given curve is rotated about the specified axis Find the volume V of the resulting solid by any method x2 y 42 16 about the yaxis class=

Respuesta :

Answer:

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]

MrRoyal

The volume of the solid is a function of its definite integral.

The volume of the solid is: [tex]\mathbf{\frac{256}{3}\pi}[/tex]

The curve is given as:

[tex]\mathbf{x^2 + (y - 4)^2 = 16}[/tex]

Solve for x^2

[tex]\mathbf{x^2 = 16 - (y - 4)^2}[/tex]

Using the washer method, we have the following integral:

[tex]\mathbf{Volume = \int\limits^b_a \pi x^2 dy}[/tex]

The graph passes through y = 0 and y = 8.

So, we have:

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

Substitute [tex]\mathbf{x^2 = 16 - (y - 4)^2}[/tex]

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

Rewrite as:

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

Expand brackets

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

Open bracket

[tex]\mathbf{Volume = \pi\int\limits^8_0 16 - y^2 + 8y - 16\ dy}[/tex]

Evaluate like terms

[tex]\mathbf{Volume = \pi\int\limits^8_0 - y^2 + 8y \ dy}[/tex]

Integrate

[tex]\mathbf{Volume = \pi( - \frac{1}{3}y^3 + 4y^2 )|\limits^8_0}[/tex]

Expand

[tex]\mathbf{Volume = \pi[( - \frac{1}{3} \times 8^3 + 4\times 8^2 ) - ( - \frac{1}{3} \times 0^3 + 4\times 0^2 )]}[/tex]

[tex]\mathbf{Volume = \pi[( - \frac{1}{3} \times 512 + 4\times 64 ) ]}[/tex]

[tex]\mathbf{Volume = \pi[( - \frac{512}{3} + 256 ) ]}[/tex]

Simplify

[tex]\mathbf{Volume = \pi[( \frac{-512 + 256 \times 3}{3}) ]}[/tex]

[tex]\mathbf{Volume = \pi[( \frac{256}{3}) ]}[/tex]

[tex]\mathbf{Volume = \frac{256}{3}\pi}[/tex]

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

Read more about volume of solids at:

https://brainly.com/question/8752196

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