Calc 2 Integration Area Problem

[Alexa]

Homework Statement

Please help me solve the calc problem pictured!

Homework Equations

y=3-x^2 and y=x+1

The Attempt at a Solution

My attempt is in one of the photos!

 

[Ray Vickson]

Homework Statement

Please help me solve the calc problem pictured!

Homework Equations

y=3-x^2 and y=x+1

The Attempt at a Solution

My attempt is in one of the photos!

Type the problem statement, and your solution. Your images are not readable on my devices, and so I am unable to help or give hints.

For more on this issue, see the post "Guidelines for students and helpers", by Vela.

 

[Alexa]

The region R is bounded by y=3−x^2 and y=x+1.
The area of the region can be found by integrating: integral from 1 to 2 ______dy + integral from 2 to 3 ______dy
For the first blank I had (sqrt(3-y))-(y-1) and for the second I had (sqrt(3-y)-2)
These are both wrong according to the system

 

[Ray Vickson]

The region R is bounded by y=3−x^2 and y=x+1.
The area of the region can be found by integrating: integral from 1 to 2 ______dy + integral from 2 to 3 ______dy
For the first blank I had (sqrt(3-y))-(y-1) and for the second I had (sqrt(3-y)-2)
These are both wrong according to the system

So, the problem statement is asking you to find the area the hard way; integrating with respect to $x$ would be a lot easier.

Anyway, to see the $x$-limits in the $y$-integrals, you should start by drawing a picture of your region. The first (vertically lower) region goes from a negative value of $y$ to a positive value, and for each such $y$, from a smaller (sometimes negative, sometimes positive) value of $x$ to a larger value of $x$---giving a positive $x$-length. You have your first integral going from a large value of $x$ to a smaller one---giving a negative $x$-length.

You seem to be assuming that $x$ must be positive, but that is not stated anywhere in the problem as you wrote it.

 

[vela]

The region R is bounded by y=3−x^2 and y=x+1.
The area of the region can be found by integrating: integral from 1 to 2 ______dy + integral from 2 to 3 ______dy
For the first blank I had (sqrt(3-y))-(y-1) and for the second I had (sqrt(3-y)-2)
These are both wrong according to the system

Could you explain to us how you came up with your attempt?