Question

The curve y = 1 -(1/4)x^2 intersects the positive side of the x-axis at A and the y-axis at B. O is the origin. Calculate the volume generated when the finite area bounded by BO, OA, and the arc AB is rotated through four right angles
i. about the x-axis
ii. about the y-axis.
Give each answer as a multiple of π.

Answer 1

FORMULAS: $V=\displaystyle\int A dx$, or respectively $\displaystyle\int Ady$ where $A$ stands for the area of the typical disc. Another words: $A = \pi r^2$ and $r=f(x)$ or $r=f(y)$ depending on the axis of revolution.

• The volume of the solid generated by a region under $f(x)$ bounded by the $x$ - axis and vertical lines $x=a$ and $x=b$, which is revolved about the x-axis is

$$V=\pi \displaystyle\int\limits_a^by^2dx=\pi\displaystyle\int\limits_a^b\Big[f(x)\Big]^2dx$$

• The volume of the solid generated by a region under $f(y)$ (to the left of $f(y)$ bounded by the $y$ - axis, and horizontal lines $y=c$ and $y=d$ which is revolved about the y-axis.

$$V=\pi \displaystyle\int\limits_c^dx^2dy=\pi\displaystyle\int\limits_c^d\Big[f(y)\Big]^2dy$$

SOLUTION: $y=1-\dfrac{x^2}{4}, ~~~x=\pm2\sqrt{1-y}~;$

$$a)~ V_x = 2 \pi \displaystyle\int_{0}^{2} \Big[f(x)\Big]^2 dx = 2\pi \displaystyle\int_{0}^{2}\Bigg[\dfrac{4-x^2}{4}\Bigg]^2dx=2\pi \cdot \dfrac{16}{15}=\dfrac{32}{15}\pi;$$ $$b)~ V_y=2\pi\displaystyle\int_{0}^{1}\Big[f(y)\Big]^2dy=2\pi\displaystyle\int_{0}^{1}\Bigg[2\sqrt{1-y}\Bigg]^2dy=2\pi\cdot2=4\pi;~~~$$ $$\text{Answer:}~V_x=\dfrac{32}{15}\pi,~V_y=4\pi.$$