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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 4)2 = 16; about the y class=

Verified Answer :

Answer:

Hence the volume of the solid is: \(\dfrac{256}{3}\pi\)

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= \(\int\limits^8_0 {\pi x^2} \, dy\)

Now we are given the equation as:

\(x^2 + (y-4)^2 = 16\)

so we get the value of \(x^2\) as:

\(x^2=16-(y-4)^2\)

Now the volume is computed as:

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

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

Which is given by putting the limits.

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

Hence the volume of the solid is: \(\dfrac{256}{3}\pi\)

MrRoyal

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

The volume of the solid is: \(\mathbf{\frac{256}{3}\pi}\)

The curve is given as:

\(\mathbf{x^2 + (y - 4)^2 = 16}\)

Solve for x^2

\(\mathbf{x^2 = 16 - (y - 4)^2}\)

Using the washer method, we have the following integral:

\(\mathbf{Volume = \int\limits^b_a \pi x^2 dy}\)

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

So, we have:

\(\mathbf{Volume = \int\limits^8_0 \pi x^2 dy}\)

Substitute \(\mathbf{x^2 = 16 - (y - 4)^2}\)

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

Rewrite as:

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

Expand brackets

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

Open bracket

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

Evaluate like terms

\(\mathbf{Volume = \pi\int\limits^8_0 - y^2 + 8y \ dy}\)

Integrate

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

Expand

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

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

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

Simplify

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

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

\(\mathbf{Volume = \frac{256}{3}\pi}\)

Hence, the volume of the solid is: \(\mathbf{\frac{256}{3}\pi}\)

Read more about volume of solids at:

https://brainly.com/question/8752196

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