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Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
\(y = 7sin(x)\)
\(y = 7cos(x)\)

\(0 <= x <= \pi /4\)
rotates about y=-1

I've tried every answer I could, but still couldn't figure it out
Please help <3

StayCurly  Sep 14, 2018
 #1
avatar+27128 
+2

Like so:

 

Alan  Sep 15, 2018
 #2
avatar+93866 
+3

I just lifted everything by one unit and it becomes

 

Find the volume formed when the region between the graphs  y=7sinx+1   and   y=7cosx+1  for  0<=x<=pi/4

is rotated around the line y=0

 

\(Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[(7cosx+1)^2-(7sinx+1)^2 \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[(49cos^2x+1+14cosx)-(49sin^2x+1+14sinx) \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[(49cos^2x-49sin^2x+14cosx-14sinx) \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[49(cos^2x-sin^2x)+14cosx-14sinx) \right]dx\\ Volume = \displaystyle \pi\int_0^{\pi/4}\;\left[49(cos2x)+14cosx-14sinx) \right]dx\\ Volume =\pi\left[ \frac{49sin2x}{2}+14sinx+14cosx \right]^{\pi/4}_0\\ Volume =\pi\left[ \frac{49}{2}+\frac{14}{\sqrt2}+\frac{14}{\sqrt2} \right]-\pi\left[ 14 \right]\\ Volume =\pi\left[ \frac{49}{2}+\frac{14\sqrt2}{2}+\frac{14\sqrt2}{2}-\frac{28}{2} \right]\\ Volume =\pi\left[ \frac{41}{2}+\frac{28\sqrt2}{2}\right]\\ Volume =\frac{\pi}{2}\left[ 41+28\sqrt2\right]\quad units^3\)

 

 

 

Here is the original graph

 

 

 

I have lifted everyting by 1 unit to make it easier to work with

 

Melody  Sep 15, 2018

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