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# Volume of revolution : IIT JAM 2018 Question Number 19

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

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

## 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"]

[/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"]

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[/et_pb_text][et_pb_post_nav in_same_term="off" _builder_version="4.0.9"][/et_pb_post_nav][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

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

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

## 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"]

[/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"]

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