10.3: Double Integrals in Polar Coordinates
- Page ID
- 144352
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- Let \(R \subset \mathbb{R}^2\) be the portion of the disk of radius \(4\) centered at the origin that lies in the first quadrant. Express region \(R\) in polar coordinates.
- Let \(R \subset \mathbb{R}^2\) be the annulus that lies between the circles of radius \(1\) and radius \(2\). Express region \(R\) in polar coordinates.
- Let \(R \subset \mathbb{R}^2\) be the region in the first quadrant bounded by \(x^2 + y^2 = 1\), \(x^2 + y^2 = 9\), \(y = x\), and \(y = 0\). Express region \(R\) in polar coordinates.
- Let \(R \subset \mathbb{R}^2\) be the region bounded by \(y = x\), \(y = 0\), \(x = 1\), and \(x = 5\). Express region \(R\) in polar coordinates.
- Let \(R \subset \mathbb{R}^2\) be the region bounded by \(y = x\) and \(y = x^2\). Express region \(R\) in polar coordinates.
- Use polar coordinates to evaluate \(\displaystyle \iint_R (x^2 + y^2)\ dA\) over \(R = \{(r, \theta)\ |\ 3 \le r \le 5,\ 0 \le \theta \le 2\pi\}\).
- Use polar coordinates to evaluate \(\displaystyle \iint_R \sqrt[3]{x^2 + y^2}\ dA\) over \(R = \left\{(r, \theta)\ \big|\ 0 \le r \le 1,\ \dfrac{\pi}{2} \le \theta \le \pi\right\}\).
- Use polar coordinates to evaluate \(\displaystyle \iint_R \arctan\left(\dfrac{y}{x}\right)\ dA\) over \(R = \left\{(r, \theta)\ \big|\ 1 \le r \le 2,\ \dfrac{\pi}{4} \le \theta \le \dfrac{\pi}{3}\right\}\).
- Evaluate \(\displaystyle \int_0^3 \int_0^\sqrt{9-y^2} (x^2 + y^2)\ dx\,dy\) by converting to polar coordinates.
- Evaluate \(\displaystyle \int_0^4 \int_{-\sqrt{16 - x^2}}^\sqrt{16 - x^2} \sin(x^2 + y^2)\ dy\,dx\) by converting to polar coordinates.
- Evaluate \(\displaystyle \int_{-4}^4 \int_{-\sqrt{16 - y^2}}^0 (2y - x)\ dx\,dy\) by converting to polar coordinates.
- Evaluate \(\displaystyle \int_0^2 \int_y^\sqrt{8-y^2} (x+ y)\ dx\,dy\) by converting to polar coordinates.
- Evaluate \(\displaystyle \int_0^a \int_{-\sqrt{a^2 - x^2}}^0 x^2y\ dy\,dx\) by converting to polar coordinates.
- Evaluate \(\displaystyle \int_0^1 \int_{x^2}^x \dfrac{1}{\sqrt{x^2 + y^2}}\ dy\,dx\) by converting to polar coordinates.