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Math  /  Calculus

QuestionConvert the integral ∫02∫−xxdydx\int_{0}^{\sqrt{2}} \int_{-x}^{x} d y d x to polar coordinates and evaluate it (use tt for θ\theta ): With a=a= □\square □\square □\square and d=d= □\square ∫02∫−xxdydx=∫ab∫cd□drdt\int_{0}^{\sqrt{2}} \int_{-x}^{x} d y d x=\int_{a}^{b} \int_{c}^{d} \square d r d t =∫ab□dt=□ab=□.\begin{aligned} = & \int_{a}^{b} \square d t \\ & =\square_{a}^{b} \\ & =\square . \end{aligned}

Studdy Solution

STEP 1

1. The given integral is a double integral over a region in the Cartesian coordinate system.
2. We will convert this integral into polar coordinates.
3. The region of integration is symmetric about the y-axis.

STEP 2

1. Identify the region of integration in Cartesian coordinates.
2. Convert the region of integration to polar coordinates.
3. Set up the integral in polar coordinates.
4. Evaluate the integral.

STEP 3

Identify the region of integration in Cartesian coordinates. The limits for y y are from −x-x to xx, and the limits for x x are from 00 to 2\sqrt{2}. This describes a region symmetric about the y-axis, bounded by the lines y=−x y = -x and y=x y = x , and extending from x=0 x = 0 to x=2 x = \sqrt{2} .

STEP 4

Convert the region of integration to polar coordinates. In polar coordinates, x=rcos⁡(t) x = r \cos(t) and y=rsin⁡(t) y = r \sin(t) . The line y=x y = x becomes tan⁡(t)=1 \tan(t) = 1 , or t=π4 t = \frac{\pi}{4} . The line y=−x y = -x becomes tan⁡(t)=−1 \tan(t) = -1 , or t=−π4 t = -\frac{\pi}{4} . The radial distance r r varies from 0 0 to 2 \sqrt{2} .

STEP 5

Set up the integral in polar coordinates. The Jacobian of the transformation from Cartesian to polar coordinates is r r . Thus, the integral becomes:
∫−π4π4∫02r dr dt\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2}} r \, dr \, dt

STEP 6

Evaluate the integral. Start with the inner integral with respect to r r :
∫02r dr=[r22]02=(2)22−022=22=1\int_{0}^{\sqrt{2}} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{\sqrt{2}} = \frac{(\sqrt{2})^2}{2} - \frac{0^2}{2} = \frac{2}{2} = 1
Now evaluate the outer integral with respect to t t :
∫−π4π41 dt=[t]−π4π4=π4−(−π4)=π4+π4=π2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 \, dt = \left[ t \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}
The value of the integral is:
π2\boxed{\frac{\pi}{2}}

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