Question

Question #faf07

Answer 1

If you draw a picture (graph the function), you'll see that the region whose area you want to calculate is made up of a rectangle with base 3 and height 1 and a quarter-circle of radius 3.

Therefore, the integral equals the sum of the areas of these two regions:

`\int_{-3}^{0}(1+sqrt(9-x^2))\ dx=3\cdot 1+\frac{1}{4}pi\cdot 3^2=3+\frac{9pi}{4}\approx 10.07.`

Answer 2

Here is some help getting the picture of the region:

The graph of `y=sqrt(9-x^2)` is the upper half (`y` is positive) of the circle: `x^2+y^2 = 9`.

So here's the graph of `y=sqrt(9-x^2)`

graph{y=sqrt(9-x^2) [-5.936, 6.554, -0.9, 5.336]}

Add `+1` to get `f(x)=1+sqrt(9-x^2)`.

graph{y=1+sqrt(9-x^2) [-5.834, 6.656, -0.75, 5.49]}

Finally, we need to restrict our attention to the second quadrant ( `-3 <= x <= 0`)

graph{y= (1+sqrt(9-x^2))*(-(sqrt(-x)^2)/x) [-5.834, 6.656, -0.75, 5.49]}

Now the area under the curve is as Bill Kinney describes it and you can read his answer to finish the problem.