anonymous
  • anonymous
The volume of the solid that results when y=6/7x^2 and y=13/7-x^2 is rotated around the x-axis.
Mathematics
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anonymous
  • anonymous
The volume of the solid that results when y=6/7x^2 and y=13/7-x^2 is rotated around the x-axis.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
If you look at the attached image, you'll see you need to find the volume that's generated by rotating the area between the two parabolas about the x-axis. If the top curve is y_2 and the bottom one, y_2, the elemental volume will be given by\[\delta V \approx \pi (y_2^2-y_1^2)\delta x\](it's just area of circle generated by top curve multiplied by some 'thickness' minus area of bottom circle multiplied by some 'thickness', to give you a net element of volume). So you'd have,\[y=\pi \int\limits_{-1}^{1}y_2^2-y_1^2dx=\pi \int\limits_{-1}^{1}(\frac{13}{7}-x^2)^2-(\frac{6}{7}x^2)^2dx\]\[=\pi \int\limits_{-1}^{1} \frac{13x^4}{49}-\frac{26}{7}x^2+\frac{169}{49}dx\]\[=\frac{3328}{735}\pi\]
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dumbcow
  • dumbcow
First graph the functions to get the bounds is it bounded by y-axis? we also need to know where they meet 6/7 x^2 = 13/7 -x^2 solve for x -> x = +- 1 this means we will integrate from 0 to 1 when this graph is rotated about x-axis it creates a donut like solid with circular cross-sections Area of each cross-section is piR1^2 - piR2^2 where R1 is outer radius and measures distance from x-axis to y-value of our upper function. upper function = 13/7 -x^2 lower function = 6/7x^2 V = integral from 0 to 1 of (piR1^2 - piR2^2) dx Factor out the pi R1^2 = (13/7 -x^2)^2 = 169/49 - 26/7x^2 +x^4 R2^2 = (6/7x^2)^2 = 36/49x^4 R1^2 - R2^2 = 13/49x^4 - 26/7x^2+169/49 Factor out 1/49 V =pi/49* integral from 0to1 of 13x^4 - (26*7)x^2+169 dx V = pi/49*[13/5 - ((26*7)/3) +169] = pi/49 *(110.933) = 2.26pi This assumes graph bounded by y-axis If not multiply volume by 2
anonymous
  • anonymous
The limits -1 and 1 are those x-values where the two functions intersect, since this defines the upper and lower bounds of your interval.

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dumbcow
  • dumbcow
oh i didnt see you were answering...your explanation looks better haha
anonymous
  • anonymous
It's okay :) I like your name...

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