- #1
jaychay
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The problem is to solve for the area R.
Can you please help me ?
I have tried to do it many times.
Thank you in advice.
skeeter said:$\displaystyle V_1 - V_2 = \pi \int_1^4 [f(x)]^2 \, dx - \pi \int_1^4 [f(x)-1]^2 \, dx = 4\pi$
$\displaystyle \int_1^4 [f(x)]^2 \, dx - \int_1^4 [f(x)]^2 - 2f(x) + 1 \, dx = 4$
$\displaystyle \cancel{\int_1^4 [f(x)]^2 \, dx} - \cancel{\int_1^4 [f(x)]^2 \, dx} + \int_1^4 2f(x) - 1 \, dx = 4$
note $\displaystyle \int_1^4 f(x) \, dx = R + 3$
can you finish?
jaychay said:https://www.physicsforums.com/attachments/10787
Can you explain to me please where did (f(x)-1)^2 come from ? and why you have to put -1 behind f(x) ?
The disk method is a mathematical technique used to find the area of a solid of revolution. It involves dividing the solid into infinitely thin disks and summing their areas to find the total volume.
The disk method is used when finding the area of a solid that is created by rotating a curve around an axis. This could be a circle, parabola, or any other curve.
The steps for using the disk method are as follows:
The disk method assumes that the solid of revolution is made up of infinitely thin disks, and that the disks are all parallel to the axis of rotation. It also assumes that the curve being rotated is continuous and does not intersect itself.
Some common mistakes when using the disk method include forgetting to square the radius in the formula for the area of a disk, using the wrong limits of integration, and not taking into account any holes or gaps in the solid of revolution.