1. **Problem statement:** Given the joint density function $$f(x,y) = k(x^2 + y^2)$$ for $$30 \leq x < 50$$ and $$30 \leq y < 50$$, and zero elsewhere, we need to find several quantities related to the random variables $X$ and $Y$. 2. **Find $k$:** Since $f(x,y)$ is a joint density function, it must integrate to 1 over the support. $$\int_{30}^{50} \int_{30}^{50} k(x^2 + y^2) \, dy \, dx = 1$$ Separate the integral: $$k \int_{30}^{50} \int_{30}^{50} (x^2 + y^2) \, dy \, dx = k \int_{30}^{50} \left( \int_{30}^{50} x^2 \, dy + \int_{30}^{50} y^2 \, dy \right) dx$$ Since $x^2$ is constant with respect to $y$: $$= k \int_{30}^{50} \left( x^2 (50-30) + \left[ \frac{y^3}{3} \right]_{30}^{50} \right) dx = k \int_{30}^{50} \left( 20 x^2 + \frac{50^3 - 30^3}{3} \right) dx$$ Calculate constants: $$\frac{50^3 - 30^3}{3} = \frac{125000 - 27000}{3} = \frac{98000}{3}$$ So: $$= k \int_{30}^{50} \left( 20 x^2 + \frac{98000}{3} \right) dx = k \left[ 20 \frac{x^3}{3} + \frac{98000}{3} x \right]_{30}^{50}$$ Calculate each term: $$20 \frac{50^3}{3} = 20 \frac{125000}{3} = \frac{2,500,000}{3}$$ $$20 \frac{30^3}{3} = 20 \frac{27000}{3} = 180,000$$ $$\frac{98000}{3} \times 50 = \frac{4,900,000}{3}$$ $$\frac{98000}{3} \times 30 = 980,000$$ So the integral evaluates to: $$k \left( \left( \frac{2,500,000}{3} + \frac{4,900,000}{3} \right) - \left( 180,000 + 980,000 \right) \right) = k \left( \frac{7,400,000}{3} - 1,160,000 \right)$$ Convert $1,160,000$ to thirds: $$1,160,000 = \frac{3,480,000}{3}$$ So: $$k \left( \frac{7,400,000 - 3,480,000}{3} \right) = k \frac{3,920,000}{3} = 1$$ Therefore: $$k = \frac{3}{3,920,000} = \frac{3}{3.92 \times 10^6} \approx 7.653 \times 10^{-7}$$ 3. **Find $P(30 < X < 40, 40 < Y < 50)$:** $$P = \int_{30}^{40} \int_{40}^{50} k(x^2 + y^2) \, dy \, dx$$ Calculate inner integral: $$\int_{40}^{50} (x^2 + y^2) \, dy = x^2 (50-40) + \left[ \frac{y^3}{3} \right]_{40}^{50} = 10 x^2 + \frac{125000 - 64000}{3} = 10 x^2 + \frac{61000}{3}$$ Outer integral: $$k \int_{30}^{40} \left( 10 x^2 + \frac{61000}{3} \right) dx = k \left[ 10 \frac{x^3}{3} + \frac{61000}{3} x \right]_{30}^{40}$$ Calculate terms: $$10 \frac{40^3}{3} = 10 \frac{64000}{3} = \frac{640,000}{3}$$ $$10 \frac{30^3}{3} = 10 \frac{27000}{3} = 90,000$$ $$\frac{61000}{3} \times 40 = \frac{2,440,000}{3}$$ $$\frac{61000}{3} \times 30 = 610,000$$ So: $$k \left( \left( \frac{640,000}{3} + \frac{2,440,000}{3} \right) - (90,000 + 610,000) \right) = k \left( \frac{3,080,000}{3} - 700,000 \right)$$ Convert $700,000$ to thirds: $$700,000 = \frac{2,100,000}{3}$$ So: $$k \left( \frac{3,080,000 - 2,100,000}{3} \right) = k \frac{980,000}{3}$$ Plug in $k$: $$P = \frac{3}{3,920,000} \times \frac{980,000}{3} = \frac{980,000}{3,920,000} = 0.25$$ 4. **Find probability both tires are underfilled:** Underfilled means pressure less than 40. $$P(X < 40, Y < 40) = \int_{30}^{40} \int_{30}^{40} k(x^2 + y^2) \, dy \, dx$$ Inner integral: $$\int_{30}^{40} (x^2 + y^2) \, dy = 10 x^2 + \frac{40^3 - 30^3}{3} = 10 x^2 + \frac{64000 - 27000}{3} = 10 x^2 + \frac{37000}{3}$$ Outer integral: $$k \int_{30}^{40} \left( 10 x^2 + \frac{37000}{3} \right) dx = k \left[ 10 \frac{x^3}{3} + \frac{37000}{3} x \right]_{30}^{40}$$ Calculate terms: $$10 \frac{40^3}{3} = \frac{640,000}{3}$$ $$10 \frac{30^3}{3} = 90,000$$ $$\frac{37000}{3} \times 40 = \frac{1,480,000}{3}$$ $$\frac{37000}{3} \times 30 = 370,000$$ So: $$k \left( \left( \frac{640,000}{3} + \frac{1,480,000}{3} \right) - (90,000 + 370,000) \right) = k \left( \frac{2,120,000}{3} - 460,000 \right)$$ Convert $460,000$ to thirds: $$460,000 = \frac{1,380,000}{3}$$ So: $$k \frac{740,000}{3} = \frac{3}{3,920,000} \times \frac{740,000}{3} = \frac{740,000}{3,920,000} \approx 0.1888$$ 5. **Find covariance $\mathrm{Cov}(X,Y)$:** $$\mathrm{Cov}(X,Y) = E[XY] - E[X]E[Y]$$ Calculate $E[X]$: $$E[X] = \int_{30}^{50} \int_{30}^{50} x f(x,y) \, dy \, dx = k \int_{30}^{50} \int_{30}^{50} x (x^2 + y^2) \, dy \, dx = k \int_{30}^{50} \int_{30}^{50} (x^3 + x y^2) \, dy \, dx$$ Inner integral: $$\int_{30}^{50} (x^3 + x y^2) \, dy = x^3 (50-30) + x \left[ \frac{y^3}{3} \right]_{30}^{50} = 20 x^3 + x \frac{125000 - 27000}{3} = 20 x^3 + x \frac{98000}{3}$$ Outer integral: $$k \int_{30}^{50} \left( 20 x^3 + \frac{98000}{3} x \right) dx = k \left[ 20 \frac{x^4}{4} + \frac{98000}{3} \frac{x^2}{2} \right]_{30}^{50} = k \left[ 5 x^4 + \frac{49000}{3} x^2 \right]_{30}^{50}$$ Calculate terms: $$5 \times 50^4 = 5 \times 6,250,000 = 31,250,000$$ $$5 \times 30^4 = 5 \times 810,000 = 4,050,000$$ $$\frac{49000}{3} \times 50^2 = \frac{49000}{3} \times 2500 = \frac{122,500,000}{3}$$ $$\frac{49000}{3} \times 30^2 = \frac{49000}{3} \times 900 = \frac{44,100,000}{3}$$ So: $$k \left( (31,250,000 + \frac{122,500,000}{3}) - (4,050,000 + \frac{44,100,000}{3}) \right) = k \left( 31,250,000 - 4,050,000 + \frac{122,500,000 - 44,100,000}{3} \right)$$ Calculate: $$31,250,000 - 4,050,000 = 27,200,000$$ $$\frac{122,500,000 - 44,100,000}{3} = \frac{78,400,000}{3}$$ So: $$k \left( 27,200,000 + \frac{78,400,000}{3} \right) = k \frac{81,600,000 + 78,400,000}{3} = k \frac{160,000,000}{3}$$ Plug in $k$: $$E[X] = \frac{3}{3,920,000} \times \frac{160,000,000}{3} = \frac{160,000,000}{3,920,000} \approx 40.82$$ By symmetry, $E[Y] = E[X] = 40.82$. 6. Calculate $E[XY]$: $$E[XY] = \int_{30}^{50} \int_{30}^{50} xy f(x,y) \, dy \, dx = k \int_{30}^{50} \int_{30}^{50} xy (x^2 + y^2) \, dy \, dx = k \int_{30}^{50} \int_{30}^{50} (x^3 y + x y^3) \, dy \, dx$$ Inner integral: $$\int_{30}^{50} (x^3 y + x y^3) \, dy = x^3 \frac{y^2}{2} + x \frac{y^4}{4} \Big|_{30}^{50} = x^3 \frac{50^2 - 30^2}{2} + x \frac{50^4 - 30^4}{4}$$ Calculate: $$50^2 - 30^2 = 2500 - 900 = 1600$$ $$50^4 - 30^4 = 6,250,000 - 810,000 = 5,440,000$$ So inner integral: $$x^3 \times 800 + x \times 1,360,000$$ Outer integral: $$k \int_{30}^{50} (800 x^3 + 1,360,000 x) dx = k \left[ 800 \frac{x^4}{4} + 1,360,000 \frac{x^2}{2} \right]_{30}^{50} = k \left[ 200 x^4 + 680,000 x^2 \right]_{30}^{50}$$ Calculate terms: $$200 \times 50^4 = 200 \times 6,250,000 = 1,250,000,000$$ $$200 \times 30^4 = 200 \times 810,000 = 162,000,000$$ $$680,000 \times 50^2 = 680,000 \times 2500 = 1,700,000,000$$ $$680,000 \times 30^2 = 680,000 \times 900 = 612,000,000$$ So: $$k \left( (1,250,000,000 + 1,700,000,000) - (162,000,000 + 612,000,000) \right) = k (2,950,000,000 - 774,000,000) = k 2,176,000,000$$ Plug in $k$: $$E[XY] = \frac{3}{3,920,000} \times 2,176,000,000 = \frac{6,528,000,000}{3,920,000} \approx 1664.29$$ 7. Calculate covariance: $$\mathrm{Cov}(X,Y) = E[XY] - E[X]E[Y] = 1664.29 - (40.82)^2 = 1664.29 - 1666.91 = -2.62$$ 8. **Find correlation coefficient $\rho$:** $$\rho = \frac{\mathrm{Cov}(X,Y)}{\sigma_X \sigma_Y}$$ Calculate $\sigma_X^2 = E[X^2] - (E[X])^2$. Calculate $E[X^2]$: $$E[X^2] = \int_{30}^{50} \int_{30}^{50} x^2 f(x,y) \, dy \, dx = k \int_{30}^{50} \int_{30}^{50} x^2 (x^2 + y^2) \, dy \, dx = k \int_{30}^{50} \int_{30}^{50} (x^4 + x^2 y^2) \, dy \, dx$$ Inner integral: $$\int_{30}^{50} (x^4 + x^2 y^2) \, dy = x^4 (50-30) + x^2 \left[ \frac{y^3}{3} \right]_{30}^{50} = 20 x^4 + x^2 \frac{98000}{3}$$ Outer integral: $$k \int_{30}^{50} \left( 20 x^4 + \frac{98000}{3} x^2 \right) dx = k \left[ 20 \frac{x^5}{5} + \frac{98000}{3} \frac{x^3}{3} \right]_{30}^{50} = k \left[ 4 x^5 + \frac{98000}{9} x^3 \right]_{30}^{50}$$ Calculate terms: $$4 \times 50^5 = 4 \times 3,125,000,000 = 12,500,000,000$$ $$4 \times 30^5 = 4 \times 243,000,000 = 972,000,000$$ $$\frac{98000}{9} \times 50^3 = \frac{98000}{9} \times 125,000 = \frac{12,250,000,000}{9} \approx 1,361,111,111$$ $$\frac{98000}{9} \times 30^3 = \frac{98000}{9} \times 27,000 = \frac{2,646,000,000}{9} \approx 294,000,000$$ So: $$k \left( (12,500,000,000 + 1,361,111,111) - (972,000,000 + 294,000,000) \right) = k (13,861,111,111 - 1,266,000,000) = k 12,595,111,111$$ Plug in $k$: $$E[X^2] = \frac{3}{3,920,000} \times 12,595,111,111 \approx 9637.5$$ Calculate variance: $$\sigma_X^2 = 9637.5 - (40.82)^2 = 9637.5 - 1666.91 = 7970.59$$ Similarly, $\sigma_Y^2 = 7970.59$ and $\sigma_X = \sigma_Y = \sqrt{7970.59} \approx 89.28$. Correlation: $$\rho = \frac{-2.62}{89.28 \times 89.28} = \frac{-2.62}{7970.59} \approx -0.00033$$ 9. **Find marginal densities:** $$f_X(x) = \int_{30}^{50} f(x,y) \, dy = k \int_{30}^{50} (x^2 + y^2) \, dy = k \left( 20 x^2 + \frac{50^3 - 30^3}{3} \right) = k \left( 20 x^2 + \frac{98000}{3} \right)$$ Similarly, $$f_Y(y) = k \left( 20 y^2 + \frac{98000}{3} \right)$$ 10. **Find conditional densities:** $$f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)} = \frac{k(x^2 + y^2)}{k \left( 20 y^2 + \frac{98000}{3} \right)} = \frac{x^2 + y^2}{20 y^2 + \frac{98000}{3}}$$ Similarly, $$f_{Y|X}(y|x) = \frac{x^2 + y^2}{20 x^2 + \frac{98000}{3}}$$ **Final answers:** - $k \approx 7.653 \times 10^{-7}$ - $P(30 < X < 40, 40 < Y < 50) = 0.25$ - $P(X < 40, Y < 40) \approx 0.1888$ - $\mathrm{Cov}(X,Y) \approx -2.62$ - Correlation coefficient $\rho \approx -0.00033$ - Marginal densities: $$f_X(x) = k \left( 20 x^2 + \frac{98000}{3} \right), \quad f_Y(y) = k \left( 20 y^2 + \frac{98000}{3} \right)$$ - Conditional densities: $$f_{X|Y}(x|y) = \frac{x^2 + y^2}{20 y^2 + \frac{98000}{3}}, \quad f_{Y|X}(y|x) = \frac{x^2 + y^2}{20 x^2 + \frac{98000}{3}}$$