# Thread: Area of a 2D surface

1. ## Area of a 2D surface

I was tutoring this morning and I was stuck in this problem shamefully...

The region bounded by y=x^8 , y=1, and revolved along y=4

My solution was taking the integral of pi*[y^(1/8)]^2 dy

I was tutoring this morning and I was stuck in this problem shamefully...

The region bounded by y=x^8 , y=1, and revolved along y=4

My solution was taking the integral of pi*[y^(1/8)]^2 dy

Have you considered shifted down the curve 4 units?
The advantage of that is that you now have,
x^8-4 revolved about y=0 and use the standard formula.
Find the intersection between y=1 and x=8 are x=+/- 1
Thus,
INTEGRAL^1_{-1} pi*(x^8-4)^2 dx

The region bounded by y = x^8 , y = 1, and revolved about y = 4

You revolved the region about the x-axis . . . . We want "washers".

. . . . . . . .
1
. . V .= .π
[(x^8 - 4)² - 3²] dx
. . . . . . .
-1