Volumes of Revolution 3

Question 1 of 3

Find the volume generated when

$x = sqrt(16-y^2)$ is rotated about the

$y$ – axis, between

$y = -4$ &

$y = 4$

1. $256pi$ cubic units
2. $128pi$ cubic units
3. $(128pi)/3$ cubic units
4. $(256pi)/3$ 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

$x$	$=$	$sqrt(16-y^2)$
$x^2$	$=$	$16-y^2$	Square both sides

Substitute

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

$V$	$=$	$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$ $x^2$ $\:dy$	Limits are $y=-4$ and $y=4$
	$=$	$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$ $16-y^2$ $\:dy$	$x^2=16-y^2$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$ $16-y^2$ $\:dy$

	$=$	$\pi \left(16\frac{y^{0+1}}{0+1} – \frac{y^{2+1}}{2+1} \right)$	Apply Power Rule

	$=$	$pi (16y-(y^3)/3)$	Simplify

Find the Definite Integral

$V$	$=$	$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$ $16-y^2$ $\:dy$

	$=$	$\pi \left[16y – \frac{y^3}{3} \right]_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$

	$=$	$\pi \left[\left(16(\color{#9a00c7}{4}) – \frac{\color{#9a00c7}{4}^3}{3}\right) – \left(16(\color{#00880A}{-4})-\frac{(\color{#00880A}{-4})^3}{3}\right)\right]$	Substitute the upper $(4)$ and lower limits $(-4)$

	$=$	$pi (64-64/3)-(-64 -(-64)/3)$	Simplify

	$=$	$pi (128/3 + 128/3)$

	$=$	$(256pi)/3$

$(256pi)/3$ cubic units

Question 2 of 3

Find the volume generated when

$y = x^2$ is rotated about the

$y$ – axis, between

$y = 1$ &

$y = 4$

1. $(15pi)/2$ cubic units
2. $(15pi)/4$ cubic units
3. $5pi$ cubic units
4. $15pi$ 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

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

Substitute

$x^2$ into the given formula and substitute the limits $y=1$ and $y=4$

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

Apply Power Rule

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

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

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

Find the Definite Integral

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

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

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

	$=$	$pi (16/2)-(1/2)$	Simplify

	$=$	$pi (15/2)$

	$=$	$(15pi)/2$

$(15pi)/2$ cubic units

Question 3 of 3

Find the volume generated by the area bounded by

$y = sqrtx$ and

$y=x^2$ when rotated about the

$y$ – axis, between

$y = 0$ &

$y = 1$

1. $3/10$ cubic units
2. $(3pi)/10$ cubic units
3. $10/3$ cubic units
4. $30pi$ cubic units

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Volumes-Disc Method

$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} (x_f^2 – x_c^2) dy$

First, find the values of $x_f^2$ and $x_c^2$ .

$y$	$=$	$x^2$
$x^2$	$=$	$y$
$x_f^2$	$=$	$y$
$y$	$=$	$sqrtx$
$y^2$	$=$	$x$
$x_c$	$=$	$y^2$
$x_c^2$	$=$	$y^4$

Substitute $x_f^2$ and $x_c^2$ into the given formula and substitute the limits $x=0$ and $x=1$

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} ($ $x_f^2$ $-$ $x_c^2$ $)\:dy$	Limits are $x=0$ and $x=1$
	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} ($ $y$ $-$ $y^4$ $)\:dy$	Substitute $x_f^2$ and $x_c^2$

Apply Power Rule

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} (y-y^4)\:dy$

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

	$=$	$pi (y^2/2 -y^5/5)$	Simplify

Find the Definite Integral

$V$	$=$	$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} (y-y^4)\:dy$

	$=$	$\pi \left[\frac{y^2}{2} – \frac {y^5}{5} \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{1}}$

	$=$	$\pi \left[\left(\frac{\color{#9a00c7}{1}^2}{2} – \frac{\color{#9a00c7}{1}^5}{5}\right) – \left(\frac{\color{#00880A}{0}^2}{2} – \frac{\color{#00880A}{0}^5}{5}\right)\right]$	Substitute the upper $(1)$ and lower limits $(0)$

	$=$	$pi [(1/2-1/5)-(0)]$	Simplify

	$=$	$pi [3/10]$

	$=$	$(3/10)pi$

	$=$	$(3pi)/10$

$(3pi)/10$ cubic units

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Volumes-Disc Method

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Volumes-Disc Method

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Volumes-Disc Method

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