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3 years ago

Find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.

Mathematics

1 answer:

kherson [118]3 years ago

7 0

Answer:

$=\frac{364\pi}{3}$ cubic units

Step-by-step explanation:

We are to find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.

f(x)=2x+1, y=0, x=0, x=4.

The picture is given as shaded region.

This is rotated about x axis

Limits for x are already given as 0 and 4

f(x) is a straight line

The solid formed would be a cone

Volume = $\pi \int\limits^a_b {(2x+1)^2} \, dx \\= \pi \int\limits^4_0 {(4x^2+4x+1)} \, dx \\=\pi [\frac{4x^3}{3} +2x^2+x]^5_0\\\\=\pi[\frac{4*4^3}{3}+2*4^2+4-0]\\=\frac{364\pi}{3}$

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Read 2 more answers

What sine function represents an amplitude of 1, a period of 2π, a horizontal shift of π, and a vertical shift of −4?

Gwar [14]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}
\\\\
-------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)$

with that template in mind, let's see

the period of 2π, well, for that, you have to do nothing, because that's sine original period

Amplitude of 1, well, that's also sine's original amplitude

horizontal shift/phase... ok, that means C/B = π, now, it doesn't say if to the left or the right, so I gather either is ok, so...let's do the right

you could use then, C = -π and B = 1, that gives you -π/1 or -π

vertical shift of -4, well, that simply means D = -4

f(θ) = sin(θ - π) - 4 <-- will do

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3 years ago