Region R is bounded by curve y=x^2 , y=1/x and the line x=1/2 . Calculate the volume of the solid formed by rotating region R through 360 degree about
a) the x-axis
b) the y-axis
My problem is where is the region R in the graph?
Region R is bounded by curve y=x^2 , y=1/x and the line x=1/2 . Calculate the volume of the solid formed by rotating region R through 360 degree about
a) the x-axis
b) the y-axis
My problem is where is the region R in the graph?
a.) Around the x-axis, we have the bounds as $\displaystyle x=\frac{1}{2}...1$ because $\displaystyle \frac{1}{x}=x^2 \implies x=1$.
Note that $\displaystyle \frac{1}{x}\geq x^2$ here.
So $\displaystyle \pi\int_{1/2}^1\frac{1}{x^2}-x^4\,dx = \pi\left[-\frac{1}{x}-\frac{x^5}{5}\right]_{1/2}^1 = -\pi\frac{6}{5}-(-\pi\frac{321}{160}) = \boxed{\frac{129\pi}{160}}$
b.) I'd break it up into two integrals. Note that the functions are now $\displaystyle x=\frac{1}{y}$ and $\displaystyle x=\sqrt{y}$
$\displaystyle x=\frac{1}{y}$ intersects $\displaystyle x=\frac{1}{2}$ at $\displaystyle y=2$
$\displaystyle x=\frac{1}{y}$ intersects $\displaystyle x=\sqrt{y}$ at $\displaystyle y=1$
$\displaystyle x=\sqrt{y}$ intersects $\displaystyle x=\frac{1}{2}$ at $\displaystyle y=\frac{1}{4}$
So we have $\displaystyle \pi\int_{1/4}^1 (\sqrt{y})^2-(1/2)^2\,dy + \pi\int_1^2(1/y)^2-(1/2)^2\,dy = \frac{9\pi}{32}+\frac{\pi}{4} = \boxed{\frac{17\pi}{32}}$