rotate the region bounded by y = 6 - 2x - x^2 and y = x+6 about the line y=3

Question

3psilon · Answer

My answer differs from that of my teacher's. Can someone just show me the integral

zepdrix · Answer

Can you provide the teacher's answer a sec? I wanna see if I made some terrible mistake before I post this integral :P

3psilon · Answer

ok ok

3psilon · Answer

attachment

3psilon · Answer

Problem 21 part D

zepdrix · Answer

Hmm this is setup really strangly...
Clearly the parabola is our upper function.

So why is the parabola being subtracted from the line...
This is what I came up with,
$$\large \pi \int\limits_{-3}^0 \left[(6-2x-x^2)-3\right]^2-\left[(x+6)-3\right]^2\;dx$$

I should look over my work again though.. bit of a tricky problem. I may have made a mistake somewhere.

3psilon · Answer

But I thought the outer radius was at a height 3 plus the parabola function so why wouldn't it be 3+(6-2x-x^2)?

zepdrix · Answer

|dw:1362721421835:dw|So the outer radius $\large R$ is going to be $\large y_1-3$.
Where $\large y_1=6-2x-x^2$.