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GenaCL600 [577]
3 years ago
12

A solid lies between planes perpendicular to the x-axis at x tox=. The cross sections perpendicular to the x-axis are circular d

isks with diameters running from the curve y cotx to the curve y csc x.
Find the volume of the solid.
A. +1) OA 6 2 O
B. (2/3-2)-3 n2 O
C. (3-1)x+ (3-1)
D. 2 12
Mathematics
1 answer:
cluponka [151]3 years ago
3 0

It's unclear what the planes are supposed to be, so I'll take x=a and x=b with 0\le a.

The cross sections are disks with diameter \csc x-\cot x, so each disk of thickness \Delta x has a volume of

\dfrac{\pi(\csc x-\cot x)^2}4\Delta x

Then taking infinitesimally thin disks, we find the solid has a volume of

\displaystyle\frac\pi4\int_a^b(\csc x-\cot x)^2\,\mathrm dx

Since

(\csc x-\cot x)^2=2\csc^2x-2\csc x\cot x-1

and

\dfrac{\mathrm d(\csc x)}{\mathrm dx}=-\csc x\cot x

\dfrac{\mathrm d(\cot x)}{\mathrm dx}=-\csc^2x

it follows that the volume is

\displaystyle\frac\pi4\left(-2\cot x+2\csc x-x\right)\bigg|_a^b

=\dfrac\pi4(2(\cot a-\cot b)+2(\csc b-\csc a)+a-b)

=\dfrac\pi4\left(2\tan\dfrac b2-2\tan\dfrac a2+a-b\right)

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