Respuesta :
[tex]25x^2+49y^2=1[/tex]
Notice that if
[tex]x=\dfrac15\cos t[/tex]
[tex]y=\dfrac17\sin t[/tex]
then
[tex]\implies25\left(\dfrac15\cos t\right)^2+49\left(\dfrac17\sin t\right)^2=1[/tex]
[tex]\implies \cos^2t+\sin^2t=1[/tex]
which suggests a change of variables of
[tex]\begin{cases}x(r,t)=\dfrac r5\cos t\\\\y(r,t)\dfrac r7\sin t\end{cases}[/tex]
where [tex]0\le r\le1[/tex] and [tex]0\le t\le2\pi[/tex]. Then we also have
[tex]25x^2+49y^2=r^2[/tex]
So
[tex]\displaystyle2\iint_{\mathcal R}\sin(25x^2+49y^2)\,\mathrm dA=2\int_0^{2\pi}\int_0^1\sin(r^2)\,r\,\mathrm dr\,\mathrm dt=4\pi\int_0^1r\sin(r^2)\,\mathrm dr[/tex]
Now a replacement of [tex]r\to r^2[/tex] will finish this. We get
[tex]\displaystyle2\pi\int_0^1(2r)\sin(r^2)\,\mathrm dr=2\pi\int_0^1\sin(r^2)\,\mathrm d(r^2)[/tex]
[tex]=-2\pi\cos(r^2)\bigg|_{r^2=0}^{r^2=1}=-2\pi(\cos(1)-\cos(0))=2\pi(1-\cos(1))[/tex]
Notice that if
[tex]x=\dfrac15\cos t[/tex]
[tex]y=\dfrac17\sin t[/tex]
then
[tex]\implies25\left(\dfrac15\cos t\right)^2+49\left(\dfrac17\sin t\right)^2=1[/tex]
[tex]\implies \cos^2t+\sin^2t=1[/tex]
which suggests a change of variables of
[tex]\begin{cases}x(r,t)=\dfrac r5\cos t\\\\y(r,t)\dfrac r7\sin t\end{cases}[/tex]
where [tex]0\le r\le1[/tex] and [tex]0\le t\le2\pi[/tex]. Then we also have
[tex]25x^2+49y^2=r^2[/tex]
So
[tex]\displaystyle2\iint_{\mathcal R}\sin(25x^2+49y^2)\,\mathrm dA=2\int_0^{2\pi}\int_0^1\sin(r^2)\,r\,\mathrm dr\,\mathrm dt=4\pi\int_0^1r\sin(r^2)\,\mathrm dr[/tex]
Now a replacement of [tex]r\to r^2[/tex] will finish this. We get
[tex]\displaystyle2\pi\int_0^1(2r)\sin(r^2)\,\mathrm dr=2\pi\int_0^1\sin(r^2)\,\mathrm d(r^2)[/tex]
[tex]=-2\pi\cos(r^2)\bigg|_{r^2=0}^{r^2=1}=-2\pi(\cos(1)-\cos(0))=2\pi(1-\cos(1))[/tex]
The area bounded by the sin (25x² + 49y²) dA is calculated by using integration is negative 2π (1 - cos 1).
What is an area bounded by the curve?
When the two curves intersect then they bound the region is known as the area bounded by the curve.
Evaluate the integral by making an appropriate change of variables.
The equation of ellipse is 25x² + 49y² = 1
Let cos t = 5x and sin t = 7y. Then we have
[tex]25 (\dfrac{1}{5} \cos t)^2 + 49(\dfrac{1}{7} \sin t)^2 = 1\\\\\\\cos ^2 t + \sin^ 2 t = 1[/tex]
Which suggests a change of variable will be
[tex]\left\{\begin{matrix}x(r,t)=\dfrac{r}{5}\cos t \\\\y(r,t)=\dfrac{r}{5}\sin t\end{matrix}\right.[/tex]
Where 0 ≤ r ≤ 1 and 0 ≤ t ≤ 2π. Then we also have
25x² + 49y² = r²
Then
[tex]\rm \rightarrow 2 \int \int _R \sin (25x^2 + 49y^2) dA \\\\\rightarrow 2 \int _{0}^{2\pi} \int _{0}^{1} \sin r^2 dr \ dt \\\\\rightarrow 4\pi \int _0^1 r \sin r^2 dr[/tex]
Now r² is replaced by r, then we get
[tex]\rightarrow 2\pi \int _0^1 2r \sin r^2 \ dr \\\\\\\rightarrow 2\pi \int _0^2 \sin r^2 d(r^2 )\\\\\\\rightarrow -2\pi \cos r^2 |_{r^2=0}^{r^2=1}\\\\\\\rightarrow -2\pi (1-\cos 1)[/tex]
More about the area bounded by the curve link is given below.
https://brainly.com/question/24563834
