Answer to Question #281329 in Mechanical Engineering for Nick

Question

Evaluate the integral of 3 dx dy. Use inner limits=y^2 to 9, outer limits 0 to 3.

Determine the integral of xy dx dy. Using limits of x=0 to (y+1) and y=0 to 1.

Find the integral of (2-x-y) dx dy. Using limits of x=0 to (1-y)^0.5, and y=0 to 1.

Answer 1

a)

$$\displaystyle\int_{0}^3\displaystyle\int_{y^2}^{9}3dxdy=3\displaystyle\int_{0}^3[x]\begin{matrix} 9 \\ y^2 \end{matrix}dy$$

$$=3\displaystyle\int_{0}^3(9-y^2)dy=[27y-y^3]\begin{matrix} 3 \\ 0 \end{matrix}$$

$$=81-27-0=54$$

b)

$$\displaystyle\int_{0}^1\displaystyle\int_{0}^{y+1}xydxdy=\displaystyle\int_{0}^{1}y[\dfrac{x^2}{2}]\begin{matrix} y+1 \\ 0 \end{matrix}dy$$

$$=\dfrac{1}{2}\displaystyle\int_{0}^1(y^3+2y^2+y)dy=\dfrac{1}{2}[\dfrac{y^4}{4}+\dfrac{2y^3}{3}+\dfrac{y^2}{2}]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=\dfrac{1}{2}(\dfrac{1}{4}+\dfrac{2}{3}+\dfrac{1}{2})=\dfrac{17}{24}$$

c)

$$\displaystyle\int_{0}^1\displaystyle\int_{0}^{\sqrt{1-y}}(2-x-y)dxdy$$

$$=\displaystyle\int_{0}^{1}[2x-\dfrac{x^2}{2}-xy]\begin{matrix} \sqrt{1-y} \\ 0 \end{matrix}dy$$

$$=\displaystyle\int_{0}^{1}(2\sqrt{1-y}-\dfrac{1-y}{2}-y\sqrt{1-y})dy$$

$$=[-\dfrac{4(1-y)^{3/2}}{3}-\dfrac{y}{2}+\dfrac{y^2}{4}-\dfrac{2(1-y)^{5/2}}{5}+\dfrac{2(1-y)^{3/2}}{3}]\begin{matrix} 1 \\ 0 \end{matrix}$$

$$=-\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{2}{3}+\dfrac{2}{5}=\dfrac{49}{60}$$