Disc Method?

[sammay]

I used the disc method to find the volume for this problem:
y=10/(x^2), x=1, x=5
where it is revolving about the region bounded by the graph about the line y=10.

I used the disc method and this was my setup.
Volume= pi * (definite integration of [10^2 - (10/x^2)^2]dx from 1 to 5)

But my answer was 137,600 pi/ 375, which does't seem correct. Can someone explain my mistake? Should I use the shell method instead?

 

[pappus]

I used the disc method to find the volume for this problem:
y=10/(x^2), x=1, x=5
where it is revolving about the region bounded by the graph about the line y=10.

I used the disc method and this was my setup.
Volume= pi * (definite integration of [10^2 - (10/x^2)^2]dx from 1 to 5)

But my answer was 137,600 pi/ 375, which does't seem correct. Can someone explain my mistake? Should I use the shell method instead?

1. The radius of the disc is calculated by:

\(\displaystyle r = 10 - \frac{10}{x^2}\)

The volume of one disc is \(\displaystyle V_{single\ disc} = \pi \cdot r^2 \cdot dx\)

2. The volume of the solid is the sum of all discs:

\(\displaystyle V = \pi \int_1^5\left(10-\frac{10}{x^2}\right)^2 dx = \pi \int_1^5\left(100-2 \cdot 10 \cdot \frac{10}{x^2} + \frac{100}{x^4}\right) dx\)

3. Using This integral I've got \(\displaystyle V = \frac{4096}{15} \cdot \pi\)