promotion image of download ymail app
Promoted

Calculus II Shell Method?

I am learning how to use the shell method, and I get it for the most part, but this question is stumping me because I'm not sure what to put as my interval.

Attached is an image of the problem so you can see the graph if needed. I am asked to revolve y = 1 - x about the x-axis using shell method. I know that I should put it in terms of y (x = y - 1) and the interval is the graph with relations to it's y-position, but I'm not sure since it goes negative do I do [-2,0], [0,-2], [0,2], etc. for my interval? Any help is greatly appreciated, thanks!

Update:

Meant to type x = 1 - y

Attachment image

1 Answer

Relevance
  • 8 months ago
    Favourite answer

    You're integrating w.r.t. y, so express x in terms of y:

    x = 1 - y

    At any particular y value (with y<0 in the blue shaded triangle), the width of the shaded rectangle is 3 - x = 3 - (1 - y) = 2 + y and the height is |dy|. The radius (distance from the axis of revolution) is |y|, so the element of volume swept out by revolving that rectangle about the x-axis is:

    dV = 2πrh dr = 2π |y| (2 + y) |dy|

    The absolute value bars are in the way. For y<0, you can replace |y| with -y and by integrating in the positive direction (from y=-2 to 0) you make dy positive and can drop the bars:

    V = 2π ∫ (-y)(2 + y) dy = -2π ∫ (y² + 2y) dy . . . . integral from y=-2 to 0

    You can handle that integral, right? That shows working the problem "in place". Honestly, I'd probably have flipped the thing about the x-axis, with y = x - 1 as the line and moving the "blue triangle" to the 1st quadrant. I make fewer mistakes without so many minus signs in the problem.

    • Commenter avatarLog in to reply to the answers
Still have questions? Get answers by asking now.