4. Evaluate $$ \iint_{R}\left(x^{2}-x y+y^{2}\right) d A $$, where $$ R $$ is the region bounded by the ellipse $$ x^{2}-x y+y^{2}=2 $$ by using the transformation $$ x=\sqrt{2} u-\sqrt{2 / 3} v, \quad y=\sqrt{2} u+\sqrt{2 / 3} v $$

Question

UpStudy AI · Accepted Answer

We start with the integral

$$
I=\iint_{R}\left(x^{2}-xy+y^{2}\right)\,dA,
$$

where $$ R $$ is the region bounded by the ellipse

$$
x^{2}-xy+y^{2}= 2.
$$

A transformation is given by

$$
x=\sqrt{2}\,u-\sqrt{\frac{2}{3}}\,v,\quad y=\sqrt{2}\,u+\sqrt{\frac{2}{3}}\,v.
$$

We'll follow these steps:

---

**1. Expressing the Quadratic Form in $$(u,v)$$:**

Define

$$
A=\sqrt{2}\,u \quad \text{and} \quad B=\sqrt{\frac{2}{3}}\,v.
$$

Then

$$
x=A-B,\quad y=A+B.
$$

Compute
$$
\begin{aligned}
x^2 &= (A-B)^2 = A^2-2AB+B^2,\[1mm]
y^2 &= (A+B)^2 = A^2+2AB+B^2,\[1mm]
xy &= (A-B)(A+B)= A^2-B^2.
\end{aligned}
$$

Substitute these into the quadratic form:

$$
\begin{aligned}
x^2-xy+y^2 &= \Bigl[A^2-2AB+B^2\Bigr] - \Bigl[A^2-B^2\Bigr] + \Bigl[A^2+2AB+B^2\Bigr]\[1mm]
&= \left(A^2-2AB+B^2 -A^2+B^2+A^2+2AB+B^2\right)\[1mm]
&= A^2 + 3B^2.
\end{aligned}
$$

Recall that

$$
A^2 = 2u^2 \quad \text{and} \quad B^2=\frac{2}{3}v^2.
$$

Thus,

$$
x^2-xy+y^2 = 2u^2+ 3\left(\frac{2}{3}v^2\right)= 2u^2+2v^2= 2\left(u^2+v^2\right).
$$

---

**2. Transforming the Region:**

The original region is defined by

$$
x^2-xy+y^2=2.
$$

Using the new variables, this becomes

$$
2\left(u^2+v^2\right)=2 \quad \Longrightarrow \quad u^2+v^2=1.
$$

Thus the region transforms into the unit disk in the $$(u,v)$$-plane.

---

**3. Computing the Jacobian:**

The transformation is

$$
x=\sqrt{2}\,u-\sqrt{\frac{2}{3}}\,v,\quad y=\sqrt{2}\,u+\sqrt{\frac{2}{3}}\,v.
$$

The Jacobian matrix is

$$
J=\begin{pmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\[1mm]
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{pmatrix}=
\begin{pmatrix}
\sqrt{2} & -\sqrt{\frac{2}{3}}\[1mm]
\sqrt{2} & \sqrt{\frac{2}{3}}
\end{pmatrix}.
$$

Its determinant is

$$
\begin{aligned}
\det(J)&=\sqrt{2}\cdot\sqrt{\frac{2}{3}} - \left(-\sqrt{\frac{2}{3}}\cdot\sqrt{2}\right)\[1mm]
&=\sqrt{2}\sqrt{\frac{2}{3}}+\sqrt{2}\sqrt{\frac{2}{3}} =2\sqrt{2}\sqrt{\frac{2}{3}}.
\end{aligned}
$$

Since

$$
\sqrt{2}\sqrt{\frac{2}{3}}=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}},
$$

the determinant becomes

$$
\det(J)=2\cdot\frac{2}{\sqrt{3}}=\frac{4}{\sqrt{3}}.
$$

Thus, the area element transforms as

$$
dA = \left|\det(J)\right|\,du\,dv=\frac{4}{\sqrt{3}}\,du\,dv.
$$

---

**4. Rewriting the Integral:**

The integrand becomes

$$
x^2-xy+y^2=2\left(u^2+v^2\right),
$$

so the integral is

$$
\begin{aligned}
I &=\iint_{R} \left(x^2-xy+y^2\right)\,dA\[1mm]
&=\iint_{u^2+v^2\le 1} 2\left(u^2+v^2\right)\cdot\frac{4}{\sqrt{3}}\,du\,dv\[1mm]
&=\frac{8}{\sqrt{3}} \iint_{u^2+v^2\le 1}\left(u^2+v^2\right)\,du\,dv.
\end{aligned}
$$

---

**5. Converting to Polar Coordinates:**

Let

$$
u=r\cos\theta,\quad v=r\sin\theta,
$$

with $$r\in [0,1]$$ and $$\theta\in [0,2\pi]$$. The area element becomes $$du\,dv = r\,dr\,d\theta$$ and

$$
u^2+v^2=r^2.
$$

Thus the integral becomes

$$
\begin{aligned}
I &=\frac{8}{\sqrt{3}} \int_{0}^{2\pi}\int_{0}^{1} r^2\cdot r\,dr\,d\theta\[1mm]
&=\frac{8}{\sqrt{3}}\int_{0}^{2\pi}\int_{0}^{1} r^3\,dr\,d\theta.
\end{aligned}
$$

Evaluate the $$r$$-integral:

$$
\int_{0}^{1} r^3\,dr=\left.\frac{r^4}{4}\right|_{0}^{1}=\frac{1}{4}.
$$

Then the $$\theta$$-integral:

$$
\int_{0}^{2\pi} d\theta=2\pi.
$$

So we have

$$
I=\frac{8}{\sqrt{3}}\cdot\frac{1}{4}\cdot2\pi = \frac{8\cdot2\pi}{4\sqrt{3}} = \frac{16\pi}{4\sqrt{3}} = \frac{4\pi}{\sqrt{3}}.
$$

---

**Final Answer:**

$$
\boxed{\frac{4\pi}{\sqrt{3}}}
$$
The value of the integral is $$ \frac{4\pi}{\sqrt{3}} $$.

Evaluate , where is the region bounded by the ellipse
by using the transformation

Solución de inteligencia artificial de Upstudy

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Calculus Questions

Evaluate , where is the region bounded by the ellipse by using the transformation