Find the volume when the region bounded by the curves $y = \sqrt{2x}$ and $y = \sqrt{4 - x}$ is rotated about the x-axis.

Neetesh Kumar | October 08, 2024

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This is the solution to Volume of Solids after Rotation
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Step-by-step solution:

To find the volume of the solid obtained by rotating the region between two curves about the x-axis, we use the formula for the volume of revolution:

$V = \pi \int_{a}^{b} \left( f(x)^2 - g(x)^2 \right) dx$

Where:

• $f(x)$ is the upper curve,
• $g(x)$ is the lower curve,
• $a$ and $b$ are the bounds of integration.

Here, the two functions are: $f(x) = \sqrt{4 - x}, \quad g(x) = \sqrt{2x}$

We need to find the points where the curves intersect to determine the limits of integration. We set the curves equal to each other to find the intersection points:

$\sqrt{4 - x} = \sqrt{2x}$

Squaring both sides: $4 - x = 2x$

$4 = 3x$

$x = \frac{4}{3}$

Thus, the region is bounded between $x = 0$ and $x = \frac{4}{3}$.

Now, the volume is given by: $V = \pi \int_{0}^{\frac{4}{3}} \left( \left( \sqrt{4 - x} \right)^2 - \left( \sqrt{2x} \right)^2 \right) dx$

Simplifying: $V = \pi \int_{0}^{\frac{4}{3}} \left( (4 - x) - (2x) \right) dx$

$V = \pi \int_{0}^{\frac{4}{3}} \left( 4 - 3x \right) dx$

Now, integrate: $V = \pi \left[ 4x - \frac{3}{2}x^2 \right]_{0}^{\frac{4}{3}}$

Substitute the limits: $V = \pi \left[ 4 \left( \frac{4}{3} \right) - \frac{3}{2} \left( \frac{4}{3} \right)^2 \right]$

$V = \pi \left[ \frac{16}{3} - \frac{3}{2} \cdot \frac{16}{9} \right]$

$V = \pi \left[ \frac{16}{3} - \frac{8}{3} \right]$

$V = \pi \cdot \frac{8}{3}$

$V = \frac{8\pi}{3}$

Final Answer:

Thus, the volume is: $\boxed{ \frac{8\pi}{3} }$

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