Why does my shell method give a different volume than the washer method?

[Donald_G]

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[EvelynAG]

I’m working through a calculus problem where I need to find the volume of a solid formed by rotating a region around the y-axis. I set up my integral using the shell method, but when I check the answer key, they used the washer method instead. I’m not sure why my approach gives a different result, and I’m wondering if I’m misidentifying the radius or height of a cylindrical shell.

[KevinMP]

Around the y axis the radius of a typical shell is the distance from the line x=0, so r equals |x|. If your region is entirely to the right of the y axis you can drop the absolute value and use r = x. If parts of the region sit on both sides you should split the integral at x = 0.

[Kyle_T]

Height in that setup is the vertical length of the region at that x, meaning h(x) = y_top(x) − y_bottom(x). If you accidentally used a horizontal distance or mixed up which curves bound the top and bottom, you’ll get the wrong numbers.

[EleanorZB]

When you switch to the washer view you’re slicing perpendicular to the axis, so you must express x as functions of y. Then outer radius is the rightmost x-value and inner radius is the leftmost x-value for each y in the region. If the region touches the axis, the inner radius is zero.

[Jack_S]

Double-check the x-range you integrate over and whether the top/bottom switches between pieces. A lot of errors come from forgetting to split at the point where the bounding curve changes, which changes h or R.

[Ryan.L]

Could you share the exact region and your two integral setups? A quick look might reveal where the mismatch lies.

[Sofia.L]

I remember doing this and getting a different number once because I forgot the radius isn’t y or something, it’s the horizontal distance. Small sign mistake or mixing up left vs right boundary sneaks in and the numbers don't match the key.