Math  /  Calculus

QuestionWhante D2dA\int_{D} 2 d A, where D={(x,y):0x1,xy1+x)D=\{(x, y): 0 \leq x \leq 1, x \leq y \leq 1+x). A.0A .0 B. 3 0.4
3. Folluate D10dA\iint_{D} 10 d A, where D={(x,y):0y2,0xy}D=\{(x, y): 0 \leq y \leq 2,0 \leq x \leq y\}. A. 1 B. 5 C. 10 D. 20
14. If f(x,y)=4f(x, y)=4, evaluate Rf(x,y)dA\iint_{R} f(x, y) d A, where R={(r,θ):1r2,0θπ}R=\{(r, \theta): 1 \leq r \leq 2,0 \leq \theta \leq \pi\}. A. π\pi B. 3π3 \pi C. 6π6 \pi D. 10π10 \pi
15. Express aa0a2x2(x2+y2)3/2dydx\int_{-a}^{a} \int_{0}^{\sqrt{a^{2}-x^{2}}}\left(x^{2}+y^{2}\right)^{3 / 2} d y d x in polar coordinates. A. 0π/20ar5drdθ\int_{0}^{\pi / 2} \int_{0}^{a} r^{5} d r d \theta B. 0π0ar4drdθ\int_{0}^{\pi} \int_{0}^{a} r^{4} d r d \theta C. 0π0ar3drdθ\int_{0}^{\pi} \int_{0}^{a} r^{3} d r d \theta D. 0π02cosθr2drdθ\int_{0}^{\pi} \int_{0}^{2 \cos \theta} r^{2} d r d \theta
16. Express Exz3dV\iiint_{E} x z^{3} d V, where E={(x,y,z):0x1,0yx,0zx2y2}E=\left\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq x, 0 \leq z \leq x^{2} y^{2}\right\} in a form that can be evaluated. A. 0x2y20x01xz3dxdydz\int_{0}^{x^{2} y^{2}} \int_{0}^{x} \int_{0}^{1} x z^{3} d x d y d z B. 0x010x2y2xz2dzdxdy\int_{0}^{x} \int_{0}^{1} \int_{0}^{x^{2} y^{2}} x z^{2} d z d x d y C. 010x2y20xxz2dydzdx\int_{0}^{1} \int_{0}^{x^{2} y^{2}} \int_{0}^{x} x z^{2} d y d z d x D). 010x0x2y2xz2dzdydx\int_{0}^{1} \int_{0}^{x} \int_{0}^{x^{2} y^{2}} x z^{2} d z d y d x
17. | et E={(r,0,z):00π,0r2,1zr2}E=\left\{(r, 0, z): 0 \leq 0 \leq \pi, 0 \leq r \leq 2,1 \leq z \leq r^{2}\right\} and f(x,y,z)=2f(x, y, z)=2. lixpress Ef(x,y,z)dV\iiint_{E} f(x, y, z) d V in cylindrical condinates. ค. 0π021r2zrd%drd0\int_{0}^{\pi} \int_{0}^{2} \int_{1}^{r^{2}} z r d \% d r d 0
13. 011021r2%d%drd()\int_{0}^{11} \int_{0}^{2} \int_{1}^{r^{2}} \% d \% d r d() (i. 03x021r2zr2dzdrd0\int_{0}^{3 x} \int_{0}^{2} \int_{1}^{r^{2}} z r^{2} d z d r d 0 1) 1111121r2zrd%d0dr\int_{1}^{11} \int_{11}^{2} \int_{1}^{r^{2}} z r d \% d 0 d r

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Studdy Solution

STEP 1

1. The region DD is defined in the xyxy-plane.
2. The function to integrate is constant, f(x,y)=2f(x, y) = 2.

STEP 2

1. Understand the region of integration DD.
2. Set up the double integral.
3. Evaluate the integral.

STEP 3

The region DD is defined by the inequalities 0x10 \leq x \leq 1 and xy1+xx \leq y \leq 1+x. This describes a region in the xyxy-plane that is bounded by the lines y=xy = x and y=1+xy = 1 + x for xx ranging from 0 to 1.

STEP 4

Set up the double integral over the region DD:
01x1+x2dydx\int_{0}^{1} \int_{x}^{1+x} 2 \, dy \, dx

STEP 5

Evaluate the inner integral with respect to yy:
x1+x2dy=2[y]x1+x=2((1+x)x)=2\int_{x}^{1+x} 2 \, dy = 2[y]_{x}^{1+x} = 2((1+x) - x) = 2
Now, evaluate the outer integral with respect to xx:
012dx=2[x]01=2(10)=2\int_{0}^{1} 2 \, dx = 2[x]_{0}^{1} = 2(1 - 0) = 2
The value of the integral is:
2 \boxed{2}

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