Hi. I have a few calculus problems that I have to do using the washers and shell method to find out what the volume of the graph rotated is. I'll post one question at a time with my work. The problem is, I should be getting the same answer for the problem with the shell method and the washer method, but I'm getting two different answers. If someone could check my work, it would be GREATLY appreciated! And please keep explanations in as easy a form as you can. I'm having a really hard time understanding these concepts. THANKS!!!

1. Find the volume of the solid that results when the region bounded by y=x and y=x^2 from x=0 and x=1, is resolved about the x axis. Apply both the shell and washer method to solve.

My work:
SHELL:
h=y
r=sqrt(y)

2pi(INT from 0 to 1)y*sqrt(y)dy
2pi(INT from 0 to 1)y^3/2dy
2pi[2/5y^5/2]from 0 to 1

4pi/5 ANSWER

WASHER METHOD:
R=x
r=x^2

pi(INT from 0 to 1)x^2-x^4 dx
pi[1/3x^3 - 1/5x^5]from 0 to 1
pi[5/15 - 3/15]

2pi/15 ANSWER

I don't know how to do shell method, but, I'd say do it on your calculator and see which answer is right and go from there

Your second answer is right. In the first one, the hight is y but the radius is (sqrt(y)-y). Look at a graph to see why. So:
2pi(INT from 0 to 1)y(sqrt(y)-y)dy
2pi(INT from 0 to 1)[y^(3/2)-y^2]dy
2pi[(2/5)y^(5/2)-(1/3)y^3]
=2pi/5 Answer

In the first part of your work, when using the shell method, you correctly identified the height of each shell as y and the radius as sqrt(y). However, there seems to be a small mistake in setting up the integral. The correct integral should be:

2π * (integral from 0 to 1) [y * (sqrt(y) - y)] dy

Expanding this integral further:

2π * (integral from 0 to 1) (y^1.5 - y^2) dy

Now, integrating:

2π * [(2/5)y^(2.5) - (1/3)y^3] evaluated from 0 to 1

Simplifying further:

2π * [((2/5)*(1)^(2.5) - (1/3)*(1)^3) - ((2/5)*(0)^(2.5) - (1/3)*(0)^3)]

2π * [(2/5) - (1/3)]

= 2π * [(6/15) - (5/15)]
= 2π * (1/15)
= 2π/15

So the volume obtained using the shell method is 2π/15.

As for the washer method, your calculations are correct, and you obtained the volume as 2π/15.

Therefore, the volumes obtained using both methods are the same, 2π/15.

If you're having trouble visualizing or understanding these concepts, I would recommend trying to plot the graph and visualize the rotation. You can also consider using online graphing tools or calculators to help you visualize the problem.