## pdd21 one year 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.

1. pdd21

@campbell_st

2. campbell_st

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

hope this helps

4. pdd21

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

Do i solve it the same?

6. campbell_st

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

so, $V=\pi \int\limits_{0}^{9} (x^2)$ I don't seem to be getting the correct answer ;/ @campbell_st

8. pdd21

nvm! I caught my mistake! haha

9. pdd21

Thankk you!(:

10. campbell_st

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

oops forgot the y^2... $V = \pi \int\limits_{0}^{9} y dy = \pi [\frac{y^2}{2}]^9_{0}$

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