anonymous 4 years ago Can someone please help me with this? What is the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis? y= sq root(3x) y=3, x=0

1. anonymous

maybe x=3 not y=3

2. anonymous

no it is y=3

3. anonymous

$\int\limits_{0}^{3}3*\pi*xdx=1.5x^2*\pi|=1.5*9*\pi$ i dont know how to solve if there y=3

4. dumbcow

|dw:1328690052250:dw| $V = \pi \int\limits_{0}^{3}R^{2} -r^{2} dx$ $V = \pi \int\limits_{0}^{3}3^{2} -(\sqrt{3x})^{2} dx$

5. anonymous

Thank you very much!

6. anonymous

Can you help me with one more? Same problem but y= 4x-x^2, y=x about the y-axis

7. dumbcow

sure find where they intersect x = 4x-x^2 x(x-3) = 0 --> x = 0 and 3 |dw:1328691017992:dw| For this one i will use the shell method so we don't have to get the function in terms of y radius = x height = top function - bottom function $V = 2\pi \int\limits_{0}^{3}x((4x-x^{2}) -x) dx$

8. anonymous

Thanks buddy!

9. dumbcow

yw