Calculus Area Problem: Integrating with respect x or y axis

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width.

y = x+1 , y = 11-x^2 , x=-1 and 2

graphic
http://www.wolframalpha.com/input/?i=y+%…

Integrate with respect to x

2 points and 2 equations to combine

A = integral (-1 to 2) [ 11 - x^2 - (x + 1) ] dx
A = integral (-1 to 2) [ - x^2 - x + 10 ] dx

Integrating with respect to y would require at least 6 points and
their single equation for each integral.

integral (-1 to 2) [ - x^2 - x + 10 ] dx = 25.5

y = x+1
y = 11-x^2
x + 1 = 11 -x^2
x^2 + x - 10 = 0
x = 1/2*(-1 - √41) => y = 1/2*(1 - √41)
x = 1/2*(-1 + √41) => y = 1/2*(1 + √41)

A1 = ∫ (-x^2 - x + 10) dx from 1/2*(-1 - √41) to 1/2*(-1 + √41)
= -x^3/3 -x^2/2 + 10x from 1/2*(-1 - √41) to 1/2*(-1 + √41)
= 1/12* [41√41 - 61) - (- 41√41 - 61)]
= 1/6*(41√41)
≈ 43.75 units^2

Edit: Sorry missed the limits x= -1 & x = 2, then area would be A1 - A2 as follow:
A2 = ∫ (-x^2 - x + 10) dx from -1 to 2 = 51/2

A = 43.75 - 51/2 = 18.25 units^2

Integrate with respect to x axis.
The bounds would be from 2 to -1
And the inside of the integral would look like 11-x^2-x-1 or 10-x^2-x
You should be able to do the rest.

x axis. (int x + 1) - (int 11 - x ^ 2) : [-1, 2] , then take the absolute value