# Thread: Application of integrate- Volume 2

1. ## Application of integrate- Volume 2

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?

2. Originally Posted by sanikui
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 $x=\frac{1}{2}...1$ because $\frac{1}{x}=x^2 \implies x=1$.

Note that $\frac{1}{x}\geq x^2$ here.

So $\pi\int_{1/2}^1\frac{1}{x^2}-x^4\,dx = \pi\left[-\frac{1}{x}-\frac{x^5}{5}\right]_{1/2}^1 = -\pi\frac{6}{5}-(-\pi\frac{321}{160}) = \boxed{\frac{129\pi}{160}}$

b.) I'd break it up into two integrals. Note that the functions are now $x=\frac{1}{y}$ and $x=\sqrt{y}$

$x=\frac{1}{y}$ intersects $x=\frac{1}{2}$ at $y=2$

$x=\frac{1}{y}$ intersects $x=\sqrt{y}$ at $y=1$

$x=\sqrt{y}$ intersects $x=\frac{1}{2}$ at $y=\frac{1}{4}$

So we have $\pi\int_{1/4}^1 (\sqrt{y})^2-(1/2)^2\,dy + \pi\int_1^2(1/y)^2-(1/2)^2\,dy = \frac{9\pi}{32}+\frac{\pi}{4} = \boxed{\frac{17\pi}{32}}$

3. Part a) i can get your mean
Part b) i can't
In part b):
intersects at
Is'nt y=1/4 ?

4. Thanks...i done.
In part b), y = 1/4
Then my answer will get 1.668971097
Thanks.

5. Originally Posted by sanikui
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?
see graph ...

6. Yeah, it should be $\frac{1}{4}$. I fixed my above post. My only excuse is that it was late at night when I answered you, but still unexcusable. Sorry if I made you waste your time.

7. No problem.
It's ok.
Thanks.