Answer :
To evaluate the integral [tex]\(\iint_R (4x - y) \, dA\)[/tex], where [tex]\(R\)[/tex] is the region in the first quadrant enclosed by the circle [tex]\(x^2+y^2=4\)[/tex] and the lines [tex]\(x=0\)[/tex] and [tex]\(y=x\)[/tex], we will change to polar coordinates. Let us proceed step-by-step:
1. Understand the Region [tex]\(R\)[/tex]:
The region [tex]\(R\)[/tex] is in the first quadrant, bounded by:
- The circle given by [tex]\(x^2 + y^2 = 4\)[/tex],
- The line [tex]\(x = 0\)[/tex] (the [tex]\(y\)[/tex]-axis),
- The line [tex]\(y = x\)[/tex].
2. Set Up Polar Coordinates:
Polar coordinates are defined by:
[tex]\[ x = r \cos \theta, \quad y = r \sin \theta, \][/tex]
with the Jacobian determinant [tex]\(dA = r \, dr \, d\theta\)[/tex].
3. Determine the Bounds:
- For the radius [tex]\(r\)[/tex], it ranges from 0 to the boundary of the circle, which is [tex]\(r = 2\)[/tex].
- For the angle [tex]\(\theta\)[/tex], it starts from the [tex]\(y\)[/tex]-axis ([tex]\(\theta = 0\)[/tex]) and goes to the line [tex]\(y = x\)[/tex] ([tex]\(\theta = \frac{\pi}{4}\)[/tex]).
Therefore, the bounds are:
[tex]\[ 0 \leq r \leq 2, \quad 0 \leq \theta \leq \frac{\pi}{4}. \][/tex]
4. Transform the Integrand:
Convert the integrand [tex]\(4x - y\)[/tex] using polar coordinates:
[tex]\[ 4x - y = 4(r \cos \theta) - (r \sin \theta) = r (4 \cos \theta - \sin \theta). \][/tex]
5. Set Up the Integral:
Incorporate the Jacobian [tex]\(r\)[/tex] into the integrand:
[tex]\[ \iint_R (4x - y) \, dA = \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} r (4 \cos \theta - \sin \theta) \, r \, dr \, d\theta. \][/tex]
Simplify the integrand:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} r^2 (4 \cos \theta - \sin \theta) \, dr \, d\theta. \][/tex]
6. Evaluate the Integral:
First, integrate with respect to [tex]\(r\)[/tex]:
[tex]\[ \int_{0}^{2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}. \][/tex]
Substitute this result back into the integral:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \frac{8}{3} (4 \cos \theta - \sin \theta) \, d\theta. \][/tex]
Factor out the constant [tex]\(\frac{8}{3}\)[/tex]:
[tex]\[ \frac{8}{3} \int_{0}^{\frac{\pi}{4}} (4 \cos \theta - \sin \theta) \, d\theta. \][/tex]
Now, integrate with respect to [tex]\(\theta\)[/tex]:
[tex]\[ \int_{0}^{\frac{\pi}{4}} 4 \cos \theta \, d\theta - \int_{0}^{\frac{\pi}{4}} \sin \theta \, d\theta. \][/tex]
The integral of [tex]\(4 \cos \theta\)[/tex] is:
[tex]\[ 4 \left[\sin \theta \right]_0^{\frac{\pi}{4}} = 4 \left( \sin \frac{\pi}{4} - \sin 0 \right) = 4 \left( \frac{\sqrt{2}}{2} - 0 \right) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}. \][/tex]
The integral of [tex]\(-\sin \theta\)[/tex] is:
[tex]\[ - \left[-\cos \theta \right]_0^{\frac{\pi}{4}} = \cos 0 - \cos \frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}. \][/tex]
Combine these two results:
[tex]\[ 2\sqrt{2} - \frac{2 - \sqrt{2}}{2}. \][/tex]
Simplify the expression inside the integral:
[tex]\[ 2\sqrt{2} - \frac{2 - \sqrt{2}}{2} = 2\sqrt{2} - 1 + \frac{\sqrt{2}}{2}. \][/tex]
Finally, multiply the integral result by [tex]\(\frac{8}{3}\)[/tex]:
[tex]\[ \frac{8}{3} \left[ 2\sqrt{2} - 1 + \frac{\sqrt{2}}{2} \right]= \frac{8}{3} \cdot (2\sqrt{2} - 1 + \frac{\sqrt{2}}{2}). \][/tex]
Combining and simplifying:
\[
\frac{8}{3} \left(2\sqrt{2} + \frac{\sqrt{2}}{2} - 1\right).
\ جل
This involves simplifying further to the final indirect integral through here, which results to the precise:
[tex]\(2 \sqrt{2} \pi-1\)[/tex] multiplication through complete to complete polar coordinates.
Thus, the result of the integral [tex]\(\iint_R (4x - y) \, dA\)[/tex] is:
\[
2 \sqrt{2} \pi-1\)
1. Understand the Region [tex]\(R\)[/tex]:
The region [tex]\(R\)[/tex] is in the first quadrant, bounded by:
- The circle given by [tex]\(x^2 + y^2 = 4\)[/tex],
- The line [tex]\(x = 0\)[/tex] (the [tex]\(y\)[/tex]-axis),
- The line [tex]\(y = x\)[/tex].
2. Set Up Polar Coordinates:
Polar coordinates are defined by:
[tex]\[ x = r \cos \theta, \quad y = r \sin \theta, \][/tex]
with the Jacobian determinant [tex]\(dA = r \, dr \, d\theta\)[/tex].
3. Determine the Bounds:
- For the radius [tex]\(r\)[/tex], it ranges from 0 to the boundary of the circle, which is [tex]\(r = 2\)[/tex].
- For the angle [tex]\(\theta\)[/tex], it starts from the [tex]\(y\)[/tex]-axis ([tex]\(\theta = 0\)[/tex]) and goes to the line [tex]\(y = x\)[/tex] ([tex]\(\theta = \frac{\pi}{4}\)[/tex]).
Therefore, the bounds are:
[tex]\[ 0 \leq r \leq 2, \quad 0 \leq \theta \leq \frac{\pi}{4}. \][/tex]
4. Transform the Integrand:
Convert the integrand [tex]\(4x - y\)[/tex] using polar coordinates:
[tex]\[ 4x - y = 4(r \cos \theta) - (r \sin \theta) = r (4 \cos \theta - \sin \theta). \][/tex]
5. Set Up the Integral:
Incorporate the Jacobian [tex]\(r\)[/tex] into the integrand:
[tex]\[ \iint_R (4x - y) \, dA = \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} r (4 \cos \theta - \sin \theta) \, r \, dr \, d\theta. \][/tex]
Simplify the integrand:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \int_{0}^{2} r^2 (4 \cos \theta - \sin \theta) \, dr \, d\theta. \][/tex]
6. Evaluate the Integral:
First, integrate with respect to [tex]\(r\)[/tex]:
[tex]\[ \int_{0}^{2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}. \][/tex]
Substitute this result back into the integral:
[tex]\[ \int_{0}^{\frac{\pi}{4}} \frac{8}{3} (4 \cos \theta - \sin \theta) \, d\theta. \][/tex]
Factor out the constant [tex]\(\frac{8}{3}\)[/tex]:
[tex]\[ \frac{8}{3} \int_{0}^{\frac{\pi}{4}} (4 \cos \theta - \sin \theta) \, d\theta. \][/tex]
Now, integrate with respect to [tex]\(\theta\)[/tex]:
[tex]\[ \int_{0}^{\frac{\pi}{4}} 4 \cos \theta \, d\theta - \int_{0}^{\frac{\pi}{4}} \sin \theta \, d\theta. \][/tex]
The integral of [tex]\(4 \cos \theta\)[/tex] is:
[tex]\[ 4 \left[\sin \theta \right]_0^{\frac{\pi}{4}} = 4 \left( \sin \frac{\pi}{4} - \sin 0 \right) = 4 \left( \frac{\sqrt{2}}{2} - 0 \right) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}. \][/tex]
The integral of [tex]\(-\sin \theta\)[/tex] is:
[tex]\[ - \left[-\cos \theta \right]_0^{\frac{\pi}{4}} = \cos 0 - \cos \frac{\pi}{4} = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}. \][/tex]
Combine these two results:
[tex]\[ 2\sqrt{2} - \frac{2 - \sqrt{2}}{2}. \][/tex]
Simplify the expression inside the integral:
[tex]\[ 2\sqrt{2} - \frac{2 - \sqrt{2}}{2} = 2\sqrt{2} - 1 + \frac{\sqrt{2}}{2}. \][/tex]
Finally, multiply the integral result by [tex]\(\frac{8}{3}\)[/tex]:
[tex]\[ \frac{8}{3} \left[ 2\sqrt{2} - 1 + \frac{\sqrt{2}}{2} \right]= \frac{8}{3} \cdot (2\sqrt{2} - 1 + \frac{\sqrt{2}}{2}). \][/tex]
Combining and simplifying:
\[
\frac{8}{3} \left(2\sqrt{2} + \frac{\sqrt{2}}{2} - 1\right).
\ جل
This involves simplifying further to the final indirect integral through here, which results to the precise:
[tex]\(2 \sqrt{2} \pi-1\)[/tex] multiplication through complete to complete polar coordinates.
Thus, the result of the integral [tex]\(\iint_R (4x - y) \, dA\)[/tex] is:
\[
2 \sqrt{2} \pi-1\)
