- #1

fishturtle1

- 394

- 82

## Homework Statement

Problem 1: Use double integrals to find the volume of the solid obtained by the rotation of the region:

##\triangle = \left\{ (x, y, z) | x^2 \le z \le 6 - x, 0 \le x \le 2, y = 0 \right\} ## (edit) in the xz-plane about the z axis

## Homework Equations

Volume = ##\int_a^b \int_c^d f(r, \theta) r dr d\theta##

## The Attempt at a Solution

When I draw this solid, I get a cone who's base is a circle with radius 4.

The height, z, is from z = 4 to z = 6. The sides of the cone are the line 6-x, rotated around the z axis.

##D = \left\{ (r, \theta)_p | r: 0 \rightarrow 4, \theta: 0 \rightarrow 2\pi \right\}##

##f(r, \theta) = 6 - r\cos(\theta)##

V = ##\int_0^{2\pi} \int_0^4 6 - r\cos(\theta) rdrd\theta##

##=\int_0^{2\pi} \int_0^4 6r - r^2\cos(\theta) drd\theta##

##=\int_0^{2\pi} 2r^2 - \frac {r^3}{3}\cos(\theta) d\theta## ...evaluate (r = 4) - (r = 0).

when i typed \right|_0^4 it didn't work..

##= \int_0^{2\pi} 48 - \frac {64}{3}\cos(\theta) d\theta##

##= 48\theta - \frac {64}{3}\sin(\theta)## ...plug in ##\theta = 2\pi## into the expression and subtract ##\theta = 0##.

##= 96\pi + 0 - (0+0)##

##= 96\pi##

but this answer looks way to big for the solid I drew.. where did I first go wrong?

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