Whats the area of a region in the first quadrant between the graph of #y= xsqrt(4-x^2)# and the x axis?

Question

Charles Arocha · Accepted Answer

The area is #A=8/3#.

The domain of the function #y(x)# graph{xsqrt(4-x^2) [-10, 10, -5, 5]} is: #(4-x^2) > 0# or #-2 <=x <= 2# so the area between the graph and the x axis in the first quadrant is: #A=int_0^2 xsqrt(4-x^2)dx# Substitute #x=2sint# #A=int_0^(pi/2) 2sintsqrt(4-4sin^2t)*2costdt# #A=8 int_0^(pi/2) sintcostsqrt(cos^2t)*dt# In the interval #[0,pi/2]# #cost >0#, so #sqrt(cos^2t) = cost# #A=8 int_0^(pi/2) sintcos^2tdt=-8int_0^(pi/2) (cost )^2d(cost)# #A=-8/3 (cos^3 t) |_(t=0)^(t=pi/2)=8/3#

Emma Aldridge · Answer

undefined To find the area of the region in the first quadrant between the graph of $ y = x \sqrt{4 - x^2} $ and the x-axis, you need to compute the definite integral of the function from the x-coordinate where it intersects the x-axis to the x-coordinate where it intersects the y-axis. The area can be calculated as follows:

$$ A = \int_{a}^{b} y \, dx $$

Where $ a $ and $ b $ are the x-coordinates of the points of intersection between the curve and the x-axis. To find these points, set $ y = 0 $ and solve for $ x $. After finding $ a $ and $ b $, integrate $ y $ with respect to $ x $ from $ a $ to $ b $ to get the area.

Isabella Ackerman · Answer

undefined To find the area of the region in the first quadrant between the graph of $y = x\sqrt{4 - x^2}$ and the x-axis, you can use definite integration.

1. First, determine the bounds of integration. Since we're interested in the first quadrant, where $x$ is positive, the bounds of integration will be the x-values where the function intersects the x-axis. Set $y = 0$ and solve for $x$ to find these points.

$$0 = x\sqrt{4 - x^2}$$

2. Solve for $x$:
$$0 = x\sqrt{4 - x^2}$$
$$0 = x(2 - x)(2 + x)$$

This equation yields three solutions: $x = 0$, $x = 2$, and $x = -2$. Since we're interested in the first quadrant, we take $x = 0$ and $x = 2$ as the bounds of integration.

3. Set up the integral to find the area:
$$A = \int_{0}^{2} x\sqrt{4 - x^2} \, dx$$

4. Integrate the function with respect to $x$ over the interval $[0, 2]$.

5. Evaluate the definite integral:
$$A = \int_{0}^{2} x\sqrt{4 - x^2} \, dx$$

After evaluating this integral, you'll find the area of the region in the first quadrant between the graph of $y = x\sqrt{4 - x^2}$ and the x-axis.