Easy Intergration (1 Viewer)

scardizzle · Jul 24, 2009

Some Math in Focus Q's

i've completely forgotten how to find the volume of a solid of revolution formed between 2 lines

1. Find the volume of the solid of revolution formed if the area enclosed between y= x^2 and y= (x-2)^2 is rotated about the x axis

2. The area bounded by the curve y = x^2 and the line y = x+2 is rotated about the x axis. Find the exact volume of the solid formed

jet · Jul 24, 2009

scardizzle · Jul 24, 2009

jetblack2007 said:
$\text{This is actually 2 volumes; the area under } x^2 \text{ between 0 and 1, and the area under } (x - 2)^2 \text{ between 1 and 2} \\ \text{In general, } V = \pi \int^a_b y^2 \, dx \\ \therefore V = \pi \int^1_0 x^4 \, dx + \pi\int^2_1 (x - 2)^4 \, dx \\ = \pi \int^1_0 x^4 \, dx + \pi\int^2_1 x^4 - 8x^3 + 24x^2 - 32x + 16 \, dx \\ = \pi \left [\frac{x^5}{5} \right ]^1_0 + \pi \left [\frac{x^5}{5} - 2x^4 + 8x^3 - 16x^2 + 16x \right ]^2_1 \\ = \frac{\pi}{5} + \pi \left [\frac{32}{5} - 32 + 64 - 64 + 32 - \frac{1}{5} + 2 - 8 + 16 - 16 \right ] \\ = \frac{\pi}{5} + \frac{\pi}{5} \\ = \frac{2\pi}{5}$

$\text{Solving } y = x^2 \text{ and } y = x + 2 \text{ simultaneously will give us our limits} \\ x^2 - x - 2= 0 \\ (x - 2)(x + 1) = 0 \\ x = -1 \text{ OR } x = 2 \\ \text{Thus, we integrate between }-1 \text{ and } 2 \\ \text{We use the volume formula above, but subtract the squares of the radii, knowing that } y = x + 2\text{ is above } y=x^2 \text{ in this domain.} \\ \therefore V = \pi \int^2_{-1} (x + 2)^2 - x^4 \, dx \\ = \pi \int^{2}_{-1} x^2 + 4x + 4 - x^4 \, dx \\ = \pi \left [\frac{x^3}{3} + 2x^2 + 4x - \frac{x^5}{5} \right ]^{2}_{-1} \\ = \pi \left [\frac{8}{3} + 8 + 8 - \frac{32}{5} + \frac{1}{3} - 2 + 4 - \frac{1}{5} \right ] \\ = \frac{72\pi}{15}$

Thanks man I get it now

Easy Intergration (1 Viewer)

scardizzle

Salve!

jet

Banned

scardizzle

Salve!

Users Who Are Viewing This Thread (Users: 0, Guests: 1)