Find the volume of the solid that results when the region enclosed by the given curves is revolved about the..

Question

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis. (Use "pi" in your exact answer)

y=sqrt(100-x^2), y=6

Thanks! :)

Answer 1

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis. (Use "pi" in your exact answer)

y = √(100 - x²), y=6

When y = 6, 100 - x² = 36

So x² = 64

Whence x = -8, 8

So required volume = ∫[from x = -8 to 8] π (y1² - y2²) dx

= π∫[from x = -8 to 8] (100 - x²) - 6²)dx

= π∫[from x = -8 to 8] (64 - x²)dx

= π[64x - ⅓x³][from x = -8 to 8]

= π[(512 - 512/3) - (-512 + 512/3)]

= 2048π/3 units³

Answer 2

this isn't too demanding. in case you draw the diagram you'll see that it really is a very good common one, y starts off from 7 at x = 0 to 7.2^(-a million/4) at x = pi/4 and is going right down to 0 at x = pi/2. So we only want V = pi. int(pi/4 to pi/2) y^2 dx = pi. int(pi/4 to pi/2) 40 9 cos x dx = pi. 40 9 sin x [pi/4 to pi/2] = 40 9 pi [sin pi/2 - sin pi/4] = 40 9 pi (a million - a million/sqrt(2)) in case you opt to rationalise the denominator you could rewrite this as = 40 9 pi (a million - sqrt(2) / 2).