Volume of revolution : IIT JAM 2018 Question Number 19

Competency in Focus: Application of Calculus (Volume of Revolution)
This problem is from IIT JAM 2018 (Question number 19) and is based on the calculation of the volume of revolution.

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 the 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} $

Do you really need a hint? Try it first!

Hint 1

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 straight 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.

Hint 2

Calculate the volumes of solid of revolution

Let's construct a narrow rectangle of base width $\mathrm d x$ and height $f(x)$ sitting under the curve. When 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.

HINT 3

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

HINT 4

Now in the given problem replace (x) by (z) , in the above 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
$V=\pi \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}[(1-z^{2})-\frac{1}{4})]\mathrm d z$.
$=\pi[\frac{3z}{4}-\frac{z^{3}}{3}]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}$
$=\pi(\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{24})\times 2$
$=\pi \frac{3\sqrt{3} \times 2 \times 2}{24}$
$=\frac{\pi \sqrt{3}}{2}$

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Integral Calculus by Gorakh Prasad

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Acute angles between surfaces: IIT JAM 2018 Qn 6

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Warm yourself up with an MCQ

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0.9" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]In $latex \Bbb R^3$ the cosine of acute angle between the surfaces $latex x^2+y^2+z^2-9=0$ and $latex z-x^2-y^2+3=0$ at the point $latex (2,1,2)$ is

$latex \frac{8}{5\sqrt{21}}$
$latex \frac{10}{5\sqrt{21}}$
$latex \frac{8}{3\sqrt{21}}$
$latex \frac{10}{3\sqrt{21}}$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" custom_padding="|0px||||"][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="4.0.9" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.0.9" hover_enabled="0"]IIT JAM 2018 Qn no 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0.9" hover_enabled="0" open="off"]Multivable calculus[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0.9" hover_enabled="0" open="on"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" hover_enabled="0" open="off"]

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2

Tom M. Apostol

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Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0.9"]If we are asked to give the angle between two lines then it is very easy to calculate but our forehead will get skinned whenever we will be asked to find out the acute angle between two lines and even worse if we are asked to find the angle between two surfaces. Surprisingly it is not very hard if think stepwise. Observe when you are asked to find out the angle between two lines you calculate it in terms slope. So basically you are firing putting the gun on someone else's shoulder. Here the question is to find that shoulder when it comes in finding the angle between two curves. Observe from the conception of the intersection of two curves that the tangent line of those curves also intersects and we have their corresponding slopes. Bingo! why not calculating the acute angle of the tangent lines and call them the angle between two curves. Now can you think how to calculate the acute angle between to surfaces?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0.9"]The acute angle between two surfaces would be the acute angle between their tangent plane. You can stop here and try to do the problem by your own otherwise continue... The main idea of finding tangent planes revolves around finding gradient of the corresponding surfaces. (For more info see question no 5). Can you calculate the gradient of the surfaces $latex x^2+y^2+z^2-9$ and $latex z-x^2-y^2+3$ at $latex (2,1,2)$?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0.9"]The gradient of the surfaces $latex f=x^2+y^2+z^2-9$ and $latex g=z-x^2-y^2+3$ at $latex (2,1,2)$ are $latex n_1=f_xi +f_yj+f_zk$ and $latex n_2=g_xi +g_yj+g_zk$ at $latex (2,1,2)$ which is $latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$. Now given these two gradients, can you find out the angle between them?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0.9"]$latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$.

This follows the cosine angles between two gradient is $latex cos \theta=|\frac{n_1.n_2}{|n_1||n_2|}|=|\frac{-16-4+4}{\sqrt{36 \times 21}}|=\frac{8}{3\sqrt{21}}$

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Explanation of hints with graph

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Knowledge Graph

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