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pdd21

  • 2 years ago

Calc Problem: Help please! Find the volume of the solid obtained by rotating the region bounded by y=9 x^2, x = 1, and y = 0, about the x-axis.

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  1. pdd21
    • 2 years ago
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    @campbell_st

  2. campbell_st
    • 2 years ago
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    ok the the formula for rotating about the x axis is \[V = \pi \int\limits_{a}^{b} y^2 dx\] so the values of a and b are and then y = 0 the corresponding value of x is x = 0... so that means a = 0 and b = 1 the graph looks like |dw:1386823443784:dw| so you are looking at \[V = \pi \int\limits_{0}^{1} (9x^2)^2 dx\] so just simply the (9x^2)^2 then integrate and evaluate between x = 1 and x = 0

  3. campbell_st
    • 2 years ago
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    hope this helps

  4. pdd21
    • 2 years ago
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    Thank you so much that helped clear up my confusion! there was another problem I was stuck on, but instead of it being respect to x-axis it's respect to the y-axis. -> Find the volume of the solid obtained by rotating the region bounded by y=x^2 , y = 0, and x = 3, about the y-axis. how do I solve this problem? @campbell_st

  5. pdd21
    • 2 years ago
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    Do i solve it the same?

  6. campbell_st
    • 2 years ago
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    ok... so find the upper value of the curve at x = 3 so y = 9 this is similar to before |dw:1386823955174:dw| the values of a and b are a = 0 and b = 9 so you are looking at \[V = \pi \int\limits_{a}^{b} x^2 dy\] you have your equation in terms of x^2 so its \[V = \pi \int\limits_{0}^{9} y dy\] hope it makes sense

  7. pdd21
    • 2 years ago
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    so, \[V=\pi \int\limits_{0}^{9} (x^2)\] I don't seem to be getting the correct answer ;/ @campbell_st

  8. pdd21
    • 2 years ago
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    nvm! I caught my mistake! haha

  9. pdd21
    • 2 years ago
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    Thankk you!(:

  10. campbell_st
    • 2 years ago
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    well because you are rotating about the y axis you integrate with respect to y \[V = \pi \int\limits_{0}^{9} y dy = 2\pi[\frac{y}{2}]^9_{0}\]

  11. campbell_st
    • 2 years ago
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    oops forgot the y^2... \[V = \pi \int\limits_{0}^{9} y dy = \pi [\frac{y^2}{2}]^9_{0}\]

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