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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.

Calculus1
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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
hope this helps

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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
Do i solve it the same?
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
so, \[V=\pi \int\limits_{0}^{9} (x^2)\] I don't seem to be getting the correct answer ;/ @campbell_st
nvm! I caught my mistake! haha
Thankk you!(:
well because you are rotating about the y axis you integrate with respect to y \[V = \pi \int\limits_{0}^{9} y dy = 2\pi[\frac{y}{2}]^9_{0}\]
oops forgot the y^2... \[V = \pi \int\limits_{0}^{9} y dy = \pi [\frac{y^2}{2}]^9_{0}\]

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