Find the volume generated by revolving the parabola y^2=2ax about y-axis

Answer: The volume generated by revolving the parabola y²=2ax about y-axis is equal to (2√2πa³)/5 unit.

In this page, we study the volume generated by revolving a curve (parabola) about y-axis with formula. The formulas for the volume are given below:

Formula

• The formula for the volume V generated by revolving a curve y=f(x) from a to b about the x-axis is: 

V = π$\displaystyle \int_a^b$ [f(x)]² dx.

•  The formula for the volume V generated by revolving a curve x=f(y) from c to d around the y-axis is: 

V = π$\displaystyle \int_c^d$ [f(y)]² dy.

Question

Find the volume generated by revolving the parabola y²=2ax about y-axis bounded by x=a.

Answer:

The given parabola is y²=2ax.

Lets express the parabola in the form x=f(y).

y²=2ax

⇒ $x= \dfrac{y^2}{2a}$

Thus, we have f(y) = $\dfrac{y^2}{2a}$

As x=a, so from y2=2ax we have that y2=2a⋅a = 2a2 ⇒ y = √2a, -√2a. So y ranges from -√2a to √2a.

Therefore, the volume V obtained by revolving the parabola y²=2ax about y-axis is

V = π $\displaystyle \int_c^d [f(y)]^2~dy$

= π $\displaystyle \int_{-\sqrt{2}a}^{\sqrt{2}a} \left( \dfrac{y^2}{2a}\right)^2~dy$

= $\pi \cdot \dfrac{1}{4a^2} \displaystyle \int_{-\sqrt{2}a}^{\sqrt{2}a} y^4~dy$

= $\dfrac{\pi}{20a^2}\Big[ y^5 \Big]_{-\sqrt{2}a}^{\sqrt{2}a}$

= $\dfrac{\pi}{20a^2}\left( (\sqrt{2}a)^5 – (-\sqrt{2}a)^5 \right)$

= $\dfrac{\pi}{20a^2} \times 2 \times (\sqrt{2}a)^5$, using the fact (-a)⁵ = -a⁵.

= $\dfrac{\pi}{10a^2} \times 4\sqrt{2}a^5$

= $\dfrac{2\sqrt{2}\pi a^3}{5}$

Therefore, the volume generated by revolving the parabola y²=2ax about y-axis is $\dfrac{2\sqrt{2}\pi a^3}{5}$ unit.

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