area under a curve about y=1

[lilblevs11]

Did the (1/2) base * h to check my equation but im doing something wrong trying to figure out the correct way.

 

[lilblevs11]

By the way the answers i'm getting are 1 and 7 pi/3

 

[daon2]

The area is certainly 1/2, not 1.

The Volume, is the volume of a cone, having radius 1. \(\displaystyle V_{\text{cone}} = \pi r^2h/3\), which gives pi/3.

Your integrand is incorrect. Recall r(x) is the distance from the axis of rotation to your function. As you have it, you are revolving about the x-axis and not the line y=1.

EDIT: Actually, now I'm not sure what the question even was. Your title suggests area of a planar region, but your work suggests you are looking for the volume of a rotation (and also possibly the area too?). Or is the problem about surface area?

 

[mmm4444bot]

The phrase "area under a curve" generally means the area between the curve and the x-axis. You are considering only the area between the lines y=x and y=1 (from x=1 through x=2), yes?

Otherwise, a cylinder is possible.

It's best to type the given exercise word-for-word, so that we may see all of the given information.

Please read the posting guidelines. Cheers :cool: