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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"]

### Similar Problems

[/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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