Respuesta :
The integral expression [tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dA[/tex] is given as a polynomial function
The result of the integral [tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dA[/tex] is [tex]\frac{\pi}{140}(1 - \cos(1))[/tex]
How to evaluate the integral?
The integral expression is given as:
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dA[/tex]
To transform the ellipse in the form [tex]a^2x^2 + b^2y^2 = 1[/tex], we make use of the following equation
[tex]u = ax[/tex] and [tex]v = by[/tex]
So, we have:
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy = \int\limits \int\limits^{}_R {\sin(u^2 + v^2)|J|} \, dudv[/tex]
Where:
[tex]u^2 = 49x^2[/tex] and [tex]v^2 = 25y^2[/tex]
This gives
[tex]u = 7x[/tex] and [tex]v = 5y[/tex]
And
[tex]J = \frac{x}{u} * \frac{y}{v}[/tex]
[tex]J = \frac{x}{7x} * \frac{y}{5y}[/tex]
Evaluate the product
[tex]J = \frac{1}{35}[/tex]
Substitute [tex]J = \frac{1}{35}[/tex] in [tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy = \int\limits \int\limits^{}_R {\sin(u^2 + v^2)|J|} \, dudv[/tex]
So, we have:
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy = \int\limits \int\limits^{}_R {\sin(u^2 + v^2)|\frac{1}{35}|} \, dudv[/tex]
This gives
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy = \int\limits \int\limits^{}_R {\sin(u^2 + v^2)\frac{1}{35}} \, dudv[/tex]
Factor out 1/35
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =\frac{1}{35} \int\limits \int\limits^{}_R {\sin(u^2 + v^2)} \, dudv[/tex]
The equation is in the first quadrant.
So, we have:
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =\frac{1}{35} \int\limits^{\pi/2}_0 \int\limits^{1}_0 {\sin(r^2)} \, rdrd\theta[/tex]
Integrate
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =\frac{1}{35} * [\theta]|\limits^{\pi/2}_0 * [-\frac 12\cos(r^2)]|\limits^{1}_0[/tex]
Evaluate the product
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =-\frac{1}{70} * [\theta]|\limits^{\pi/2}_0 * [\cos(r^2)]|\limits^{1}_0[/tex]
Expand
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =-\frac{1}{70} * [\frac{\pi}{2} - 0] * [\cos(1) - \cos(0)][/tex]
Evaluate the difference
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =-\frac{1}{70} * [\frac{\pi}{2} ] * [\cos(1) - \cos(0)][/tex]
Evaluate cos(0)
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =-\frac{1}{70} * [\frac{\pi}{2} ] * [\cos(1) -1][/tex]
Evaluate the product
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =-\frac{\pi}{140}(\cos(1) -1)[/tex]
This gives
[tex]\int\limits \int\limits^{}_R {\sin(49x^2 + 25y^2)} \, dxdy =\frac{\pi}{140}(1 - \cos(1))[/tex]
Hence, when the integral is evaluated, the result is [tex]\frac{\pi}{140}(1 - \cos(1))[/tex]
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