Volumes of Revolution 1

Question 1 of 4

Find the volume generated when

y = x^{3}

$y = x^3$ is rotated about the

$x$ – axis, between

$x = 0$ &

$x = 2$

1. $(128pi)/7$ cubic units
2. $128$ cubic units
3. $128/7$ cubic units
4. $pi / 7$ cubic units

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Volumes-Disc Method

$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} y^2 dx$

First, make

$y^2$ the subject of the given equation

$y$	$=$	$x^3$
$y^2$	$=$	$x^6$

Substitute

$y^2$ into the given formula and substitute the limits $x=0$ and $x=2$

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}}$ $y^2$ $\:dx$	Limits are $x=0$ and $x=2$
	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}}$ $x^6$ $\:dx$	$y^2=x^6$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}}$ $x^6$ $\:dx$

	$=$	$\pi \left(\frac{x^{6+1}}{6+1} \right)$	Apply Power Rule

	$=$	$pi (x^7/7)$	Simplify

Find the Definite Integral

$V$	$=$	$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}} x^6 dx$

	$=$	$\pi \left[\frac{x^7}{7}\right]_{\color{#00880A}{0}}^{\color{#9a00c7}{2}}$

	$=$	$\pi \left[\frac{\color{#9a00c7}{2}^7}{7}- \frac{\color{#00880A}{0}^7}{7}\right]$	Substitute the upper $(2)$ and lower limits $(0)$

	$=$	$pi (128/7 – 0)$	Simplify

	$=$	$(128pi)/7$

$(128pi)/7$ cubic units

Question 2 of 4

Find the volume generated when

$y = 1/2x$ is rotated about the

$x$ – axis, between

$x = 0$ &

$x = 4$

1. $(16pi)/3 \text (cubic units)$
2. $(16pi)/12 \text (cubic units)$
3. $16/3 \text (cubic units)$
4. $19 \text (cubic units)$

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Volumes-Disc Method

$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} y^2 dx$

First, make

$y^2$ the subject of the given equation

$y$	$=$	$1/2x$

$y^2$	$=$	$1/4x^2$

Substitute

$y^2$ into the given formula and substitute the limits $x=0$ and $x=4$

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{4}}$ $y^2$ $\:dx$	Limits are $x=0$ and $x=4$
	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{4}}$ $\frac{1}{4}x^2$ $\:dx$	$y^2=1/4x^2$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{4}}$ $\frac{1}{4}x^2$ $\:dx$

	$=$	$\pi \left(\frac {1}{4}\cdot\frac{x^{2+1}}{2+1} \right)$	Apply Power Rule

	$=$	$1/4*(pi x^3)/3$	Simplify

	$=$	$pi (x^3/12)$

Find the Definite Integral

$V$	$=$	$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{4}} \frac {x^2}{4} dx$

	$=$	$\pi \left[\frac{x^3}{12}\right]_{\color{#00880A}{0}}^{\color{#9a00c7}{4}}$

	$=$	$\pi \left[\frac{\color{#9a00c7}{4}^3}{12}- \frac{\color{#00880A}{0}^3}{12}\right]$	Substitute the upper $(4)$ and lower limits $(0)$

	$=$	$pi (64/12 – 0)$	Simplify

	$=$	$(16pi)/3$

$(16pi)/3 \text (cubic units)$

Question 3 of 4

Find the volume generated when

$y = 2$ is rotated about the

$x$ – axis, between

$x = -3$ &

$x = 3$

1. $24pi$ cubic units
2. $6pi$ cubic units
3. $24$ cubic units
4. $4pi$ cubic units

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Volumes-Disc Method

$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} y^2 dx$

First, make

$y^2$ the subject of the given equation

$y$	$=$	$2$
$y^2$	$=$	$4$

Substitute

$y^2$ into the given formula and substitute the limits $x=-3$ and $x=3$

$V$	$=$	$\pi \int_{\color{#00880A}{-3}}^{\color{#9a00c7}{3}}$ $y^2$ $\:dx$	Limits are $x=-3$ and $x=3$
	$=$	$\pi \int_{\color{#00880A}{-3}}^{\color{#9a00c7}{3}}$ $4$ $\:dx$	$y^2=4$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{-3}}^{\color{#9a00c7}{3}}$ $4$ $\:dx$

	$=$	$4\pi \left(\frac{x^{0+1}}{0+1} \right)$	Apply Power Rule

	$=$	$4pi x$	Simplify

Find the Definite Integral

$V$	$=$	$\int_{\color{#00880A}{-3}}^{\color{#9a00c7}{3}} 4 dx$

	$=$	$4\pi \left[x\right]_{\color{#00880A}{-3}}^{\color{#9a00c7}{3}}$

	$=$	$4pi [\color{#9a00c7}{3} – (\color{#00880A}{-3})]$	Substitute the upper $(3)$ and lower limits $(-3)$

	$=$	$4pi (6)$	Simplify

	$=$	$24pi$

$24pi$ cubic units

Question 4 of 4

Find the volume generated when

$y = x^2$ is rotated about the

$x$ – axis, between

$x = 1$ &

$x = 3$

1. $242pi$ cubic units
2. $pi/5$ cubic units
3. $(242pi)/5$ cubic units
4. $242/5$ cubic units

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Volumes-Disc Method

$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} y^2 dx$

First, make

$y^2$ the subject of the given equation

$y$	$=$	$x^2$
$y^2$	$=$	$x^4$

Substitute

$y^2$ into the given formula and substitute the limits $x=1$ and $x=3$

$V$	$=$	$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$ $y^2$ $\:dx$	Limits are $x=1$ and $x=3$
	$=$	$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$ $x^4$ $\:dx$	$y^2=x^4$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$ $x^4$ $\:dx$

	$=$	$\pi \left(\frac{x^{4+1}}{4+1} \right)$	Apply Power Rule

	$=$	$pi (x^5/5)$	Simplify

Find the Definite Integral

$V$	$=$	$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$ $x^4$ $\:dx$

	$=$	$\pi \left[\frac{x^5}{5} \right]_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$

	$=$	$\pi \left[\frac{\color{#9a00c7}{3}^5}{5} – \frac{\color{#00880A}{1}^5}{5}\right]$	Substitute the upper $(3)$ and lower limits $(1)$

	$=$	$pi (243/5-1/5)$	Simplify

	$=$	$(242/5)pi$

	$=$	$(242pi)/5$

$(242pi)/5$ cubic units

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Quiz summary

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1. Question

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Well Done!

Volumes-Disc Method

2. Question

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Nice Job!

Volumes-Disc Method

3. Question

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Excellent!

Volumes-Disc Method

4. Question

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Fantastic!

Volumes-Disc Method

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