Volume of revolution : IIT JAM 2018 Question Number 19

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What are we learning ?

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Competency in Focus: Application of Calculus (Volume of Revolution)

This is problem from IIT JAM 2018 is based on calculation of volume of revolution.

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://www.cheenta.com/wp-content/uploads/2020/02/ISI_19-1.png" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="388px" height="198px" max_height="207px"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Consider the region $D$

on $yz$

plane and bounded by the line $y=\frac{1}{2}$

and the curve $y^{2}+z^{2}=1$

where $y\geq0$

. If the region $D$

is revolved about the $z-$

axis then the volume of resulting solid is $(A)\frac{\pi}{\sqrt{3}}\qquad(B)\frac{2\pi}{\sqrt{3}}$

$\qquad(C)\frac{\pi\sqrt{3}}{2}\qquad(D)\pi\sqrt{3}$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.2.2" width="100%" max_width="1024px" max_width_tablet="" max_width_phone="" max_width_last_edited="on|phone"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.2.2" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.2.2"]IIT JAM 2018, Question number 19[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.2.2" open="off"]Calculation of volume of surface revolution[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.2.2" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.2.2" open="off"]Integral Calculus by Gorakh Prasad[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]Imagine that we have a portion of a curve. $y=f(x)$

from $x=a$

to $x=b$

. In the $xy-plane$

we revolve it around a straingh line - $x$

-axis. The result is called solid of revolution Here in our next hint we will find techniques to calculate the volumes of solid of revolution.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]Let's construct a narrow rectangle of base width $\mathrm d x$

and height $f(x)$

sitting under the curve. Whent this rectangle is revolved around the $x-\text{axis}$

. We get a disk whose radius is $f(x)$

and height is $\mathrm d x$

. The volume of this disk $\mathrm d V=\pi[f(x)]^{2}\mathrm d x$

. So the total volume of the solid is $V=\int_a^b \mathrm d V=\int_a^b \pi[f(x)]^{2}\mathrm d x$

. If the curve revolved around the vertical line (such as $y$

-axis), then horizontal disks are used. If the curve can be solved for $x$

in terms of $y$

. $x=g(y)$

then the formula would be. $V=\int_a^b[g(y)]^2\mathrm d y$

. Now can you guess if the region between two curves is revolved around the axis.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]Sometimes the region between two curves is revolved around the axis and a gap is created between the solid and the axis. A rectangle within the rotated region will become a disk with a hole in it, also known as washer. If the rectangle is vertical and extends from the curve $y=g(x)$

up to the curve $y=f(x)$

, then when it is rotated around $x$

- axis, it will result in a washer with volume equal to $\mathrm d V=\pi\{[f(x)]^{2}-[g(x)]^{2}\}\mathrm d x$

. Which gives us $V=\int_a^b \pi \{[f(x)]^{2}-[g(x)]^{2}\}\mathrm d x$

.[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]Now in the given problem replace $x$

by $z$

, in the obove discussion $y=f(z)=\sqrt{1-z^{2}}, y=g(z)=\frac{1}{2}$

and the line $y=\frac{1}{2}$

intersects the curve at $(\frac{-\sqrt{3}}{2},\frac{1}{2})$

, $(\frac{\sqrt{3}}{2},\frac{1}{2})$

in terms of $(z,x)$

co-ordinate. and hence the volume is $V=\pi \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}[(1-z^{2})-\frac{1}{4})]\mathrm d z$

. $=\pi[\frac{3z}{4}-\frac{z^{3}}{3}]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}$

$=\pi(\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{24})\times 2$

$=\pi \frac{3\sqrt{3} \times 2 \times 2}{24}$

$=\frac{\pi \sqrt{3}}{2}$

[/et_pb_tab][et_pb_tab title="Explanation by images" _builder_version="4.2.2"]

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="Amc 8 Master class" url="https://www.cheenta.com/matholympiad/" url_new_window="on" image="https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.0.9" header_font="||||||||" header_text_color="#0c71c3" header_font_size="48px" body_font_size="20px" body_letter_spacing="1px" body_line_height="1.5em" link_option_url="https://www.cheenta.com/matholympiad/" link_option_url_new_window="on"]

Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/amc-8-american-mathematics-competition/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="4.0.9" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

[et_pb_section fb_built="1" _builder_version="4.2.2" width="99.8%" custom_margin="||||false|false" custom_padding="||||false|false"][et_pb_row _builder_version="4.2.2" width="100%" custom_margin="||||false|false" custom_padding="||||false|false"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="4.2.2" text_font="Aclonica|300|||||||" text_text_color="#ffffff" header_font="Aclonica|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

What are we learning ?

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]

Competency in Focus: Application of Calculus (Volume of Revolution)

This is problem from IIT JAM 2018 is based on calculation of volume of revolution.

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src="https://www.cheenta.com/wp-content/uploads/2020/02/ISI_19-1.png" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="388px" height="198px" max_height="207px"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Consider the region $D$

on $yz$

plane and bounded by the line $y=\frac{1}{2}$

and the curve $y^{2}+z^{2}=1$

where $y\geq0$

. If the region $D$

is revolved about the $z-$

axis then the volume of resulting solid is $(A)\frac{\pi}{\sqrt{3}}\qquad(B)\frac{2\pi}{\sqrt{3}}$

$\qquad(C)\frac{\pi\sqrt{3}}{2}\qquad(D)\pi\sqrt{3}$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.2.2" width="100%" max_width="1024px" max_width_tablet="" max_width_phone="" max_width_last_edited="on|phone"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.2.2" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.2.2"]IIT JAM 2018, Question number 19[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.2.2" open="off"]Calculation of volume of surface revolution[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.2.2" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.2.2" open="off"]Integral Calculus by Gorakh Prasad[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]Imagine that we have a portion of a curve. $y=f(x)$

from $x=a$

to $x=b$

. In the $xy-plane$

we revolve it around a straingh line - $x$

-axis. The result is called solid of revolution Here in our next hint we will find techniques to calculate the volumes of solid of revolution.[/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]Let's construct a narrow rectangle of base width $\mathrm d x$

and height $f(x)$

sitting under the curve. Whent this rectangle is revolved around the $x-\text{axis}$

. We get a disk whose radius is $f(x)$

and height is $\mathrm d x$

. The volume of this disk $\mathrm d V=\pi[f(x)]^{2}\mathrm d x$

. So the total volume of the solid is $V=\int_a^b \mathrm d V=\int_a^b \pi[f(x)]^{2}\mathrm d x$

. If the curve revolved around the vertical line (such as $y$

-axis), then horizontal disks are used. If the curve can be solved for $x$

in terms of $y$

. $x=g(y)$

then the formula would be. $V=\int_a^b[g(y)]^2\mathrm d y$

. Now can you guess if the region between two curves is revolved around the axis.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]Sometimes the region between two curves is revolved around the axis and a gap is created between the solid and the axis. A rectangle within the rotated region will become a disk with a hole in it, also known as washer. If the rectangle is vertical and extends from the curve $y=g(x)$

up to the curve $y=f(x)$

, then when it is rotated around $x$

- axis, it will result in a washer with volume equal to $\mathrm d V=\pi\{[f(x)]^{2}-[g(x)]^{2}\}\mathrm d x$

. Which gives us $V=\int_a^b \pi \{[f(x)]^{2}-[g(x)]^{2}\}\mathrm d x$

.[/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]Now in the given problem replace $x$

by $z$

, in the obove discussion $y=f(z)=\sqrt{1-z^{2}}, y=g(z)=\frac{1}{2}$

and the line $y=\frac{1}{2}$

intersects the curve at $(\frac{-\sqrt{3}}{2},\frac{1}{2})$

, $(\frac{\sqrt{3}}{2},\frac{1}{2})$

in terms of $(z,x)$

co-ordinate. and hence the volume is $V=\pi \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}[(1-z^{2})-\frac{1}{4})]\mathrm d z$

. $=\pi[\frac{3z}{4}-\frac{z^{3}}{3}]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}$

$=\pi(\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{24})\times 2$

$=\pi \frac{3\sqrt{3} \times 2 \times 2}{24}$

$=\frac{\pi \sqrt{3}}{2}$

[/et_pb_tab][et_pb_tab title="Explanation by images" _builder_version="4.2.2"]

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="Amc 8 Master class" url="https://www.cheenta.com/matholympiad/" url_new_window="on" image="https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.0.9" header_font="||||||||" header_text_color="#0c71c3" header_font_size="48px" body_font_size="20px" body_letter_spacing="1px" body_line_height="1.5em" link_option_url="https://www.cheenta.com/matholympiad/" link_option_url_new_window="on"]

Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/amc-8-american-mathematics-competition/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="4.0.9" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]