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How do you find the volume $V$ of the described solid $S$ where the base of $S$ is a circular disk with radius $4r$ and Parallel cross-sections perpendicular to the base are squares?

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Last updated date: 17th Jun 2024
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Answer
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Hint: Problems of this type can be easily solved by using concepts of coordinate geometry and integration. We first take the circle of the base of the described solid $S$ on the $x-y$ plane having its centre at the origin, then we take an elemental cross section parallel to $x-z$ plane in the first octant of the solid and integrating it we get the one-fourth volume of the described solid $S$ . Multiplying the volume we get to $4$ we get the entire volume of the described solid $S$ .

Complete step by step answer:
According to the problem the base of the solid $S$ is a circle
We now place the base of the solid on the $x-y$ plane in such a way that its centre stays at the origin.
We know that if a circle on $x-y$ plane has its centre at the origin with a radius $4r$ then the equation of the circle is
${{x}^{2}}+{{y}^{2}}={{\left( 4r \right)}^{2}}$
$\Rightarrow {{x}^{2}}+{{y}^{2}}=16{{r}^{2}}$
$\Rightarrow {{x}^{2}}=16{{r}^{2}}-{{y}^{2}}.....\left( \text{1} \right)$
As the square cross-sections are parallel to the $x-z$ axis we take an elemental cross section in the first octant as shown below
$\Rightarrow dV=x\cdot 2x\cdot dy$
$\Rightarrow dV=2{{x}^{2}}\cdot dy$
Substituting the value of ${{x}^{2}}$ from equation $\left( 1 \right)$ in the above equation we get
$\Rightarrow dV=2\left( 16{{r}^{2}}-{{y}^{2}} \right)\cdot dy$
To get the one-fourth volume of the solid we integrate both the sides of the above equation
$\Rightarrow \int\limits_{0}^{{{V}_{0}}}{dV}=2\int\limits_{0}^{4r}{\left( 16{{r}^{2}}-{{y}^{2}} \right)dy}$
$\Rightarrow {{V}_{0}}=2\cdot 16\int\limits_{0}^{4r}{{{r}^{2}}}dy-2\int\limits_{0}^{4r}{{{y}^{2}}}dy$
\[\Rightarrow {{V}_{0}}=32{{r}^{2}}\left[ y \right]_{0}^{4r}-\dfrac{2}{3}\left[ {{y}^{3}} \right]_{0}^{4r}\]
$\Rightarrow {{V}_{0}}=32{{r}^{2}}\left( 4r \right)-\dfrac{2}{3}{{\left( 4r \right)}^{3}}$
$\Rightarrow {{V}_{0}}=128{{r}^{3}}-\dfrac{128}{3}{{r}^{3}}$
$\Rightarrow {{V}_{0}}=\dfrac{256}{3}{{r}^{3}}$
The total volume of the solid $S$ will be $4$ times ${{V}_{0}}$
Total volume of the solid S, $V=4\cdot \dfrac{256}{3}{{r}^{3}}$
$\Rightarrow V=\dfrac{1024}{3}{{r}^{3}}$
Therefore, the volume $V$ of the described solid S is $\dfrac{1024}{3}{{r}^{3}}$ .

Note:
We must take the limits properly while doing the integration to avoid error in the value of volume. Also, we must be careful that the volume we get after the integration is just the one-fourth of the entire volume. Hence, we must multiply it with $4$ to get the actual result.