Volumes of Revolution 1

Question 1 of 4

Find the volume generated when

y = x^{3}

$y = x^3$ is rotated about the

x

$x$ – axis, between

x = 0

$x = 0$ &

x = 2

$x = 2$

1. $\frac{π}{7}$ $pi / 7$ cubic units
2. $\frac{128 π}{7}$ $(128pi)/7$ cubic units
3. $\frac{128}{7}$ $128/7$ cubic units
4. $128$ $128$ cubic units

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

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

Substitute

y^{2}

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

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

Apply Power Rule

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

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

	$=$ $=$	$π (\frac{x^{7}}{7})$ $pi (x^7/7)$	Simplify

Find the Definite Integral

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

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

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

	$=$ $=$	$π (\frac{128}{7} - 0)$ $pi (128/7 – 0)$	Simplify

	$=$ $=$	$\frac{128 π}{7}$ $(128pi)/7$

\frac{128 π}{7}

$(128pi)/7$ cubic units

Question 2 of 4

Find the volume generated when

y = \frac{1}{2} x

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

x

$x$ – axis, between

x = 0

$x = 0$ &

x = 4

$x = 4$

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

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

$y$ $y$	$=$ $=$	$\frac{1}{2} x$ $1/2x$

$y^{2}$ $y^2$	$=$ $=$	$\frac{1}{4} x^{2}$ $1/4x^2$

Substitute

y^{2}

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

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

Apply Power Rule

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

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

	$=$ $=$	$\frac{1}{4} \cdot \frac{π x^{3}}{3}$ $1/4*(pi x^3)/3$	Simplify

	$=$ $=$	$π (\frac{x^{3}}{12})$ $pi (x^3/12)$

Find the Definite Integral

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

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

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

	$=$ $=$	$π (\frac{64}{12} - 0)$ $pi (64/12 – 0)$	Simplify

	$=$ $=$	$\frac{16 π}{3}$ $(16pi)/3$

\frac{16 π}{3} cubic units

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

Question 3 of 4

Find the volume generated when

y = 2

$y = 2$ is rotated about the

x

$x$ – axis, between

x = - 3

$x = -3$ &

x = 3

$x = 3$

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

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

$y$ $y$	$=$ $=$	$2$ $2$
$y^{2}$ $y^2$	$=$ $=$	$4$ $4$

Substitute

y^{2}

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

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

Apply Power Rule

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

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

	$=$ $=$	$4 π x$ $4pi x$	Simplify

Find the Definite Integral

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

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

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

	$=$ $=$	$4 π (6)$ $4pi (6)$	Simplify

	$=$ $=$	$24 π$ $24pi$

24 π

$24pi$ cubic units

Question 4 of 4

Find the volume generated when

y = x^{2}

$y = x^2$ is rotated about the

x

$x$ – axis, between

x = 1

$x = 1$ &

x = 3

$x = 3$

1. $\frac{π}{5}$ $pi/5$ cubic units
2. $242 π$ $242pi$ cubic units
3. $\frac{242}{5}$ $242/5$ cubic units
4. $\frac{242 π}{5}$ $(242pi)/5$ cubic units

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

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

$y$ $y$	$=$ $=$	$x^{2}$ $x^2$
$y^{2}$ $y^2$	$=$ $=$	$x^{4}$ $x^4$

Substitute

y^{2}

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

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

Apply Power Rule

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

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

	$=$ $=$	$π (\frac{x^{5}}{5})$ $pi (x^5/5)$	Simplify

Find the Definite Integral

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

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

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

	$=$ $=$	$π (\frac{243}{5} - \frac{1}{5})$ $pi (243/5-1/5)$	Simplify

	$=$ $=$	$(\frac{242}{5}) π$ $(242/5)pi$

	$=$ $=$	$\frac{242 π}{5}$ $(242pi)/5$

\frac{242 π}{5}

$(242pi)/5$ cubic units

Volumes of Revolution 1

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Well Done!

Volumes-Disc Method

2. Question

Hint

Nice Job!

Volumes-Disc Method

3. Question

Hint

Excellent!

Volumes-Disc Method

4. Question

Hint

Fantastic!

Volumes-Disc Method

Quizzes