Ok so i got 92pi/3 for this question so I want to know if its right. Find the volume generated by revolving the curve bounded by y=4-x^2, the y-axis and the x-axis about the x-axis using the disk method

Question

anonymous · Answer

$$\int\limits_{0}^{2}\pi*r ^{2}dx=\int\limits_{0}^{2}\pi*(4-x ^{2})^{2}dx=\pi*[(16*2-8*2 ^{3}+2 ^{5})]=\pi*(32-64+32)=64*pi$$

anonymous · Answer

64pi is what I got

anonymous · Answer

attachment

anonymous · Answer

you are revolving this curve about the x-axis from x=0 to x=2...which bounds did you use for your integral?

anonymous · Answer

i used 0 - 4

anonymous · Answer

but the curve you are using hits the x-axis at 2, so your bounds need to be from 0 to 2

anonymous · Answer

oops i see m mistake

anonymous · Answer

thank you!

anonymous · Answer

You are welcome :-) sketch the graph i always did that when i took calc saved me major points on exams :-)

anonymous · Answer

i think i made a mistake scetching the graph. forgot to square the x