Evaluate $\iint_R X Y \sin \left(x^2+y^2\right) \, DA$ When $R=\left[0, \sqrt{\frac{7 \pi}{6}}\right] \times\left[0, \sqrt{\frac{5 \pi}{3}}\right\].

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Introduction

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In this article, we will evaluate the double integral $\iint_R x y \sin \left(x^2+y^2\right) \, dA$ where $R$ is a region in the Cartesian plane. The region $R$ is defined as the rectangle with vertices at $(0,0)$, $(\sqrt{\frac{7 \pi}{6}},0)$, $(\sqrt{\frac{7 \pi}{6}},\sqrt{\frac{5 \pi}{3}})$, and $(0,\sqrt{\frac{5 \pi}{3}})$. We will use polar coordinates to evaluate this double integral.

Polar Coordinates

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Polar coordinates are a way of representing points in the Cartesian plane using a distance from a reference point (the origin) and an angle from a reference direction (the positive x-axis). The polar coordinates of a point $(x,y)$ are given by $(r,\theta)$, where $r$ is the distance from the origin to the point and $\theta$ is the angle from the positive x-axis to the line connecting the origin to the point.

Converting to Polar Coordinates

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To convert the double integral to polar coordinates, we need to express $x$ and $y$ in terms of $r$ and $\theta$. We have $x=r\cos\theta$ and $y=r\sin\theta$. The region $R$ is a rectangle in the Cartesian plane, but it is not a simple region in polar coordinates. However, we can use the fact that the region $R$ is bounded by the curves $x=\sqrt{\frac{7 \pi}{6}}$ and $y=\sqrt{\frac{5 \pi}{3}}$ to find the limits of integration in polar coordinates.

Limits of Integration

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The region $R$ is bounded by the curves $x=\sqrt{\frac{7 \pi}{6}}$ and $y=\sqrt{\frac{5 \pi}{3}}$. In polar coordinates, the limits of integration are given by $r=\sqrt{\frac{7 \pi}{6}}$ and $\theta=0$ to $\theta=\frac{\pi}{2}$.

Evaluating the Double Integral

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Now that we have the limits of integration in polar coordinates, we can evaluate the double integral. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{\frac{7 \pi}{6}}} r^2 \sin r^2 \cos \theta \sin \theta \, dr \, d\theta$$

Simplifying the Integral

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We can simplify the integral by using the trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{\frac{7 \pi}{6}}} r^2 \sin r^2 \sin 2\theta \, dr \, d\theta$$

Evaluating the Inner Integral

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We can evaluate the inner integral by using the substitution $u=r^2$. We have:

$$\int_{0}^{\sqrt{\frac{7 \pi}{6}}} r^2 \sin r^2 \sin 2\theta \, dr = \int_{0}^{\frac{7 \pi}{6}} \sin u \sin 2\theta \, du$$

Evaluating the Outer Integral

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We can evaluate the outer integral by using the substitution $v=2\theta$. We have:

$$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{\frac{7 \pi}{6}}} r^2 \sin r^2 \sin 2\theta \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{7 \pi}{6}} \sin u \sin v \, du \, dv$$

Final Evaluation

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We can evaluate the final integral by using the trigonometric identity $\sin u \sin v = \frac{1}{2}[\cos(u-v)-\cos(u+v)]$. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{7 \pi}{6}} [\cos(u-v)-\cos(u+v)] \, du \, dv$$

Simplifying the Final Integral

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We can simplify the final integral by using the fact that the integral of $\cos(u-v)$ is $\sin(u-v)$ and the integral of $\cos(u+v)$ is $-\sin(u+v)$. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\sin(u-v)]_{0}^{\frac{7 \pi}{6}} \, dv - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [\sin(u+v)]_{0}^{\frac{7 \pi}{6}} \, dv$$

Evaluating the Final Integral

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We can evaluate the final integral by using the fact that the integral of $\sin(u-v)$ is $-\cos(u-v)$ and the integral of $\sin(u+v)$ is $-\cos(u+v)$. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \frac{1}{2} \left[-\cos\left(\frac{7 \pi}{6}-v\right)\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \left[-\cos\left(\frac{7 \pi}{6}+v\right)\right]_{0}^{\frac{\pi}{2}}$$

Final Answer

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We can simplify the final answer by using the fact that $\cos\left(\frac{7 \pi}{6}-v\right) = \cos\left(\frac{\pi}{6}+v\right)$ and $\cos\left(\frac{7 \pi}{6}+v\right) = \cos\left(\frac{\pi}{6}-v\right)$. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \frac{1}{2} \left[-\cos\left(\frac{\pi}{6}+v\right)\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \left[-\cos\left(\frac{\pi}{6}-v\right)\right]_{0}^{\frac{\pi}{2}}$$

Final Simplification

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We can simplify the final answer by using the fact that $\cos\left(\frac{\pi}{6}+v\right) = \cos\left(\frac{\pi}{6}\right)\cos(v)-\sin\left(\frac{\pi}{6}\right)\sin(v)$ and $\cos\left(\frac{\pi}{6}-v\right) = \cos\left(\frac{\pi}{6}\right)\cos(v)+\sin\left(\frac{\pi}{6}\right)\sin(v)$. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \frac{1}{2} \left[-\cos\left(\frac{\pi}{6}\right)\cos(v)+\sin\left(\frac{\pi}{6}\right)\sin(v)\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \left[-\cos\left(\frac{\pi}{6}\right)\cos(v)-\sin\left(\frac{\pi}{6}\right)\sin(v)\right]_{0}^{\frac{\pi}{2}}$$

Final Answer

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We can simplify the final answer by using the fact that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. We have:

$$\iint_R x y \sin \left(x^2+y^2\right) \, dA = \frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)+\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)-\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}}$$

Final Simplification

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# **Evaluating a Double Integral with Polar Coordinates: Q&A** ===========================================================

Introduction

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In our previous article, we evaluated the double integral $\iint_R x y \sin \left(x^2+y^2\right) \, dA$ where $R$ is a region in the Cartesian plane. The region $R$ is defined as the rectangle with vertices at $(0,0)$, $(\sqrt{\frac{7 \pi}{6}},0)$, $(\sqrt{\frac{7 \pi}{6}},\sqrt{\frac{5 \pi}{3}})$, and $(0,\sqrt{\frac{5 \pi}{3}})$. We used polar coordinates to evaluate this double integral.

Q&A

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Q: What is the region R in the Cartesian plane?

A: The region $R$ is a rectangle with vertices at $(0,0)$, $(\sqrt{\frac{7 \pi}{6}},0)$, $(\sqrt{\frac{7 \pi}{6}},\sqrt{\frac{5 \pi}{3}})$, and $(0,\sqrt{\frac{5 \pi}{3}})$.

Q: Why did we use polar coordinates to evaluate the double integral?

A: We used polar coordinates to evaluate the double integral because the region $R$ is not a simple region in Cartesian coordinates. However, in polar coordinates, the region $R$ is a simple region.

Q: What are the limits of integration in polar coordinates?

A: The limits of integration in polar coordinates are given by $r=\sqrt{\frac{7 \pi}{6}}$ and $\theta=0$ to $\theta=\frac{\pi}{2}$.

Q: How did we simplify the integral?

A: We simplified the integral by using the trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$.

Q: How did we evaluate the inner integral?

A: We evaluated the inner integral by using the substitution $u=r^2$.

Q: How did we evaluate the outer integral?

A: We evaluated the outer integral by using the substitution $v=2\theta$.

Q: What is the final answer?

A: The final answer is $\frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)+\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)-\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}}$.

Q: Can you simplify the final answer?

A: Yes, we can simplify the final answer by using the fact that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

Conclusion

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In this article, we evaluated the double integral $\iint_R x y \sin \left(x^2+y^2\right) \, dA$ where $R$ is a region in the Cartesian plane. We used polar coordinates to evaluate this double integral and simplified the integral using various trigonometric identities. The final answer is $\frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)+\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)-\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}}$.

Final Answer

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The final answer is $\boxed{\frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)+\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \left[-\frac{\sqrt{3}}{2}\cos(v)-\frac{1}{2}\sin(v)\right]_{0}^{\frac{\pi}{2}}}$.