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# Find the volume of the solid that results when the region enclosed by the given curve: $(11$$to14) 2ay2=x(x-a)2 is revolved about x-axis ,a>0 Can you answer this question? ## 1 Answer 0 votes Toolbox: • Area bounded by the curve t=f(x), the x-axis and the ordinates x=a,x=b is \int \limits_a^b f(x) dx or \int \limits _a^b y dx • If the curve lies below the x-axis for a \leq x \leq b, then the area is \int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx Step 1: 2ay^2=x(x-a)^2 is resolved about the x-axis. a > 0 The curve is symmetric about x axes, passes through the orgin, does not exist for x < 0 and a loop is formed between x=0\; and\; x=a Step 2: Volume formed by rotating the loop V= \pi \int \limits_0^a y^2 dx \quad=\large\frac{\pi}{2a}$$\int \limits_0^a x(x-a) ^2 dx$
$\quad= \large\frac{\pi}{2a} $$\int \limits_0^a (x^3-2ax^2 +a^2 x)dx \quad= \large\frac{\pi}{2a} \bigg[\frac{x^4}{4}-\frac{2ax^3}{3}+\frac{a^2x^2}{2}\bigg]_0^a \quad= \large\frac{\pi}{2a} \bigg[\frac{a^4}{4}-\frac{2ax^4}{3}+\frac{a^4}{2}\bigg] \quad= \large\frac{\pi}{24a}$$ \bigg[3a^4-8a^4+6a^4\bigg]$