[/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"]
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_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"]
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_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"]
