Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of #y=4-x^2# and #y=1+2sinx#, how do you find the area?

Question

Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of  #y=4-x^2# and #y=1+2sinx#, how do you find the area?

Ezra Adams · Accepted Answer

#R=1.764# units squared

In area problems, the first thing to do is graph the functions in question so you have a visual sense of the problem. You can check out the functions in this example graphed here. From the graph, we see that the region #R# is the curvy-looking triangle. To find the area of #R#, we subtract the area of #1+sinx# from the area of #4-x^2#. The limits of integration are from the #y#-axis #(x=0)# to the point of intersection #(x=1.102)#. Putting this in math terms, we have: #R=(int_0^1.102 4-x^2dx)-(int_0^1.102 1+2sinxdx)# The properties of integrals say we can split these two into mini-integrals like so: #R=(int_0^1.102 4dx-int_0^1.102 x^2dx)-(int_0^1.102 1dx+2int_0^1.102 sinxdx)# Now we perform the integration: #R=(4[x]_0^1.102-[x^3/3]_0^1.102)-([x]_0^1.102-2[cosx]_0^1.102)# And then evaluate: #R=(4.408-0.446)-(1.102-2(cos1.102-cos0))# #R=3.962-(1.102-2(0.452-1))# #R=3.962-(1.102+1.096)# #R=3.962-2.198# #R=1.764# units squared

Cameron Alexander · Answer

undefined To find the area of the shaded region R in the first quadrant enclosed by the y-axis and the graphs of $ y = 4 - x^2 $ and $ y = 1 + 2 \sin(x) $, follow these steps:

1. Find the points of intersection between the two curves by setting them equal to each other and solving for $ x $.
2. Integrate the difference between the upper curve and the lower curve with respect to $ x $ over the interval where they intersect.

$$ 4 - x^2 = 1 + 2 \sin(x) $$
$$ x^2 + 2 \sin(x) - 3 = 0 $$

The solutions to this equation will give the $ x $-coordinates of the points of intersection.

3. Once you have the points of intersection, integrate the function that is above the other function minus the function below it from the lower $ x $-value to the higher $ x $-value to find the area of the shaded region.

$$ \text{Area} = \int_{x_1}^{x_2} (4 - x^2) - (1 + 2 \sin(x)) \, dx $$

where $ x_1 $ and $ x_2 $ are the $ x $-coordinates of the points of intersection.