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February 2, 2020

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.

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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"]

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Similar Problems

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