Volumes of Revolution 2

Question 1 of 4

Find the volume generated when

y = x^{2} - 5 x

$y = x^2-5x$ is rotated about the

x

$x$ – axis, between

x = 0

$x = 0$ &

x = 5

$x = 5$

1. $\frac{625 π}{6}$ $(625pi)/6$ cubic units
2. $\frac{625 π}{3}$ $(625pi)/3$ cubic units
3. $\frac{625 π}{2}$ $(625pi)/2$ cubic units
4. $625 π$ $625pi$ cubic units

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Volumes-Disc Method

V = π \int_{a}^{b} y^{2} d x

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

First, make

y^{2}

$y^2$ the subject of the given equation

$y$ $y$	$=$ $=$	$x^{2} - 5 x$ $x^2-5x$
$y^{2}$ $y^2$	$=$ $=$	${(x^{2} - 5 x)}^{2}$ $(x^2-5x)^2$
$y^{2}$ $y^2$	$=$ $=$	$x^{4} - 10 x^{3} + 25 x^{2}$ $x^4-10x^3+25x^2$

Substitute

y^{2}

$y^2$ into the given formula and substitute the limits $x = 0$ $x=0$ and $x = 5$ $x=5$

$V$ $V$	$=$ $=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$ $y^2$ $\:dx$	Limits are $x=0$ and $x=5$
	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$ $x^4-10x^3+25x^2$ $\:dx$	$y^2=x^4-10x^3+25x^2$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$ $x^4-10x^3+25x^2$ $\:dx$

	$=$	$\pi \left(\frac{x^{4+1}}{4+1} -10 \frac{x^{3+1}}{3+1} + 25 \frac {x^{2+1}}{2+1} \right)$	Apply Power Rule

	$=$	$pi (x^5/5 -(10x^4)/4 + (25x^3)/3)$	Simplify

	$=$	$pi (x^5/5 -(5x^4)/2 + (25x^3)/3)$

Find the Definite Integral

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$ $x^4-10x^3+25x^2$ $\:dx$

	$=$	$\pi \left[\frac{x^5}{5} – \frac {5x^4}{2} + \frac{25x^3}{3} \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$

	$=$	$\pi \left[\left(\frac{\color{#9a00c7}{5}^5}{5}-\frac{5\cdot\color{#9a00c7}{5}^4}{2} + \frac{25\cdot\color{#9a00c7}{5}^3}{3}\right) – \left(\frac{\color{#00880A}{0}^5}{5}-\frac{5\cdot\color{#00880A}{0}^4}{2} + \frac{25\cdot\color{#00880A}{0}^3}{3}\right)\right]$	Substitute the upper $(5)$ and lower limits $(0)$

	$=$	$pi [(625-(3125)/2 +(3125)/3)-(0)]$	Simplify

	$=$	$(625/6)pi$

	$=$	$(625pi)/6$

$(625pi)/6$ cubic units

Question 2 of 4

Find the volume generated when

$y = x-1$ is rotated about the

$y$ – axis, between

$y = -1$ &

$y = 1$

1. $8$ cubic units
2. $8/3$ cubic units
3. $(8pi)/3$ cubic units
4. $24pi$ cubic units

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Volumes-Disc Method

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

First, make

$x^2$ the subject of the given equation

$y$	$=$	$x-1$
$y+1$	$=$	$x$	Add $1$ to both sides
$x$	$=$	$y+1$
$x^2$	$=$	$(y+1)^2$
$x^2$	$=$	$y^2+2y+1$

Substitute

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

$V$	$=$	$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$ $x^2$ $\:dy$	Limits are $y=-1$ and $y=1$
	$=$	$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$ $y^2+2y+1$ $\:dy$	$x^2=y^2+2y+1$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$ $y^2+2y+1$ $\:dy$

	$=$	$\pi \left(\frac{y^{2+1}}{2+1} +2 \frac{y^{1+1}}{1+1} + \frac {y^{0+1}}{0+1} \right)$	Apply Power Rule

	$=$	$pi (y^3/3 +(2y^2)/2 + y)$	Simplify

	$=$	$pi (y^3/3 +y^2 + y)$

Find the Definite Integral

$V$	$=$	$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$ $y^2+2y+1$ $\:dy$

	$=$	$\pi \left[\frac{y^3}{3} + y^2 + y \right]_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$

	$=$	$\pi \left[\left(\frac{\color{#9a00c7}{1}^3}{3} +(\color{#9a00c7}{1})^2 + \color{#9a00c7}{1}\right) – \left(\frac{(\color{#00880A}{-1})^3}{3} +(\color{#00880A}{-1})^2 + \color{#00880A}{-1}\right)\right]$	Substitute the upper $(1)$ and lower limits $(-1)$

	$=$	$pi ((1/3+2)-(-1/3))$	Simplify

	$=$	$pi (7/3-(-1/3))$

	$=$	$(8pi)/3$

$(8pi)/3$ cubic units

Question 3 of 4

Find the volume generated when

$y = x-2$ is rotated about the

$y$ – axis, between

$y = 1$ &

$y = 3$

1. $101$ cubic units
2. $102pi$ cubic units
3. $98/3$ cubic units
4. $(98pi)/3$ cubic units

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Volumes-Disc Method

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

First, make

$x^2$ the subject of the given equation

$y$	$=$	$x-2$
$y+2$	$=$	$x$	Add $2$ to both sides
$x$	$=$	$y+2$
$x^2$	$=$	$(y+2)^2$
$x^2$	$=$	$y^2+4y+4$

Substitute

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

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

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$ $y^2+4y+4$ $\:dy$

	$=$	$\pi \left(\frac{y^{2+1}}{2+1} +4 \frac{y^{1+1}}{1+1} + 4\frac {y^{0+1}}{0+1} \right)$	Apply Power Rule

	$=$	$pi (y^3/3 +(4y^2)/2 + 4y)$	Simplify

	$=$	$pi (y^3/3 +2y^2 + 4y)$

Find the Definite Integral

$V$	$=$	$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$ $y^2+4y+4$ $\:dy$

	$=$	$\pi \left[\frac{y^3}{3} + 2y^2 + 4y \right]_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$

	$=$	$\pi \left[\left(\frac{\color{#9a00c7}{3}^3}{3} +2(\color{#9a00c7}{3})^2 + 4(\color{#9a00c7}{3})\right) – \left(\frac{\color{#00880A}{1}^3}{3} +2(\color{#00880A}{1})^2 + 4(\color{#00880A}{1})\right)\right]$	Substitute the upper $(3)$ and lower limits $(1)$

	$=$	$pi ((9+18+12)-(1/3+2+4))$	Simplify

	$=$	$pi (39-(19/3))$

	$=$	$(98pi)/3$

$(98pi)/3$ cubic units

Question 4 of 4

Find the volume generated when

$y = sqrt(9-x^2)$ is rotated about the

$y$ – axis, between

$y = 0$ &

$y = 3$

1. $18pi$ cubic units
2. $pi/18$ cubic units
3. $18$ cubic units
4. $27pi$ cubic units

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Volumes-Disc Method

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

First, make

$x^2$ the subject of the given equation

$y$	$=$	$sqrt(9-x^2)$
$y^2$	$=$	$9-x^2$	Square both sides
$x^2$	$=$	$9-y^2$	Solve for $x^2$

Substitute

$x^2$ into the given formula and substitute the limits $y=0$ and $y=3$

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$ $x^2$ $\:dy$	Limits are $y=0$ and $y=3$
	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$ $9-y^2$ $\:dy$	$x^2=9-y^2$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$ $9-y^2$ $\:dy$

	$=$	$\pi \left(9\frac{y^{0+1}}{0+1} – \frac{y^{2+1}}{2+1} \right)$	Apply Power Rule

	$=$	$pi (9y-(y^3)/3)$	Simplify

Find the Definite Integral

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$ $9-y^2$ $\:dy$

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

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

	$=$	$pi [(27-9)-(0)]$	Simplify

	$=$	$pi (18)$

	$=$	$18pi$

$18pi$ cubic units

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Quiz summary

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1. Question

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Keep Going!

Volumes-Disc Method

2. Question

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Great Work!

Volumes-Disc Method

3. Question

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Well Done!

Volumes-Disc Method

4. Question

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Excellent!

Volumes-Disc Method

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