volume of solid: axes and x + y = 1 rotates about y = -1

[tsh44]

the region enclosed by the line x +y =1 and teh coordinate axes is rotated about the line y=-1. what is the volume of the solid generated?

I had the integral of (1-x+1)^2 from 0 to 1 dx times pi. I put an additonal +1 in order to compensate for the fact that it's y= -1 and not the y axis. when I did this though I got 5pi/3. The book says it should be 4pi/3. I still cannot figure out what I did wrong. Would appreciate any help.

 

[skeeter]

using the method of washers ...

\(\displaystyle \L R(x) = 1 + (1 - x) = 2 - x\)

\(\displaystyle \L r(x) = 1\)

\(\displaystyle \L V = \pi \int_0^1 (2-x)^2 - 1 dx\)

\(\displaystyle \L V = \pi \left[-\frac{(2-x)^3}{3} - x\right]_0^1\)

\(\displaystyle \L V = \pi \left[\left(-\frac{1}{3} - 1\right) - \left(-\frac{8}{3} - 0\right)\right]\)

\(\displaystyle \L V = \frac{4\pi}{3}\)

 

[tsh44]

Where did the negative in front of the (2-x)^3/ 3 in teh antiderivative come from?

 

[skeeter]

Where did the negative in front of the (2-x)^3/ 3 in teh antiderivative come from?

find the derivative of -(2-x)³/3 to answer this question.

 

[galactus]

You could use shells also.

\(\displaystyle \L\\2{\pi}\int_{0}^{1}(x+1)(1-x)dx=2{\pi}\int_{0}^{1}(1-x^{2})dx\)