Question

can you help me:
the region bounded by x=y^2 and x=2+y revolved about y=-1

Answer 1

any body help me

Answer 2

When you revolve a region around the line, you have to sum the area. A parabola or linear line usually forms a circle. The area of a circle is pi*r^2 where your r is the radius being revolved. Try to graph the equation to look at what the radius is.
You also have to find all the points that intersect to know which integrals limits to set. Your answer should be in the form of \[/int_{a}^{b}\pi*(f_1(y+1)^2-f_2(y+1)^2)dy+/int_{b}^{c}\pi*(f_1(y+1)^2-f_2(y+1)^2)dy\]

Answer 3

i need complete answer

Answer 4

\[\int_{a}^{b}{\pi*(f_1(y+1)^2-f_2(y+1)^2)dy}+\int_{b}^{c}{\pi*(f_1(y+1)^2-f_2(y+1)^2)dy}\]
There we go. Latex is so picky sometimes.

Answer 5

Mr. 2le can you complete

Answer 6

Here's a better picture. Notice how from 0 - b ( the first critical point where the two functions meet) only the function x=y^2 is revolved.
From b to c (the next point where the two functions meet) the parabola is greater than the linear function. That keys you into which one subtracts the other.

\[\int_a^b{\pi*(f_1(y+1)^2)dy}+\int_b^c{\pi*(f_1(y+1)^2-f_2(y+1)^2)dy}\]

Answer 7

i know that but i need full answer

Answer 8

If you follow my steps, you could reach that answer. If anything is unclear, you can ask questions and I will try my best to answer.

Answer 9

now i can not finf the answer because i do not know which method i should solve it



( washer method or cylinder shell

Answer 10

please show to me the answer because i want to sleep because i have an exam in chemistry

Answer 11

please