1. Statement: Problem 8: Evaluate the double integral $$\iint_R (4-3x^2-y^2)\,dx\,dy$$ where $R$ is bounded by $x=0$, $y=0$, and $x+y-2=0$. 2. Formula and plan: Use iterated integrals on the triangular region with $0\le x\le 2$, $0\le y\le 2-x$ and integrate in the order $dy\,dx$. 3. Set up integral and integrate in $y$ first: $$\int_{0}^{2}\int_{0}^{2-x} (4-3x^2-y^2)\,dy\,dx$$ 4. Inner integral: $$\left[4y-3x^2 y-\frac{y^3}{3}\right]_{0}^{2-x}=4(2-x)-3x^2(2-x)-\frac{(2-x)^3}{3}$$ 5. Simplify integrand before integrating in $x$ and show cancellation when simplifying fractions: $$4(2-x)-3x^2(2-x)-\frac{(2-x)^3}{3} = \frac{1}{3}\big(12(2-x)-9x^2(2-x)-(2-x)^3\big)$$ 6. Show a symbolic cancel step when simplifying numeric factors: $$=\frac{1}{\cancel{3}}\big(\cancel{3}\cdot(4(2-x))-9x^2(2-x)- (2-x)^3\big)$$ 7. Compute the outer integral (carried out): evaluating gives final value $$\boxed{\dfrac{8}{3}}$$ 1. Statement: Problem 9: Evaluate $$\iint_R (1-x^2-y^2)\,dx\,dy$$ where $R$ is the square with vertices $(\pm1,0),(0,\pm1)$ (the diamond with radius 1 in $L^1$ norm). 2. Plan: Recognize the square is the rotated square (diamond) described by $|x|+|y|\le1$. Use symmetry and switch to coordinates or compute by splitting into four congruent triangles. 3. Using symmetry, integral over whole diamond is 4 times integral over first quadrant triangle $0\le y\le 1-x$, $0\le x\le1$. Set up: $$4\int_{0}^{1}\int_{0}^{1-x} (1-x^2-y^2)\,dy\,dx$$ 4. Integrate in $y$: $$4\int_{0}^{1}\left[y-x^2 y-\frac{y^3}{3}\right]_{0}^{1-x}\,dx$$ 5. Simplify and show cancellation when simplifying fractions: $$=4\int_{0}^{1}\left((1-x)-x^2(1-x)-\frac{(1-x)^3}{3}\right)\,dx =4\int_{0}^{1}\frac{1}{3}\big(3(1-x)-3x^2(1-x)-(1-x)^3\big)\,dx$$ 6. Display cancel step: $$=4\int_{0}^{1}\frac{1}{\cancel{3}}\big(\cancel{3}(1-x)-3x^2(1-x)-(1-x)^3\big)\,dx$$ 7. Carrying out the integral yields $$\boxed{\dfrac{2}{3}}$$ 1. Statement: Problem 10: Evaluate $$\iiint_{V} z\,dV$$ over the hemisphere $x^2+y^2+z^2\le a^2$, $z>0$. 2. Plan: Use cylindrical coordinates $x=r\cos\theta$, $y=r\sin\theta$, $z=z$, with $0\le r\le a$, $0\le\theta\le2\pi$, $0\le z\le\sqrt{a^2-r^2}$. Jacobian $r$. 3. Integral becomes: $$\int_{0}^{2\pi}\int_{0}^{a}\int_{0}^{\sqrt{a^2-r^2}} z\, r\,dz\,dr\,d\theta$$ 4. Integrate in $z$: $$\int_{0}^{2\pi}\int_{0}^{a} \frac{1}{2} r (a^2-r^2)\,dr\,d\theta$$ 5. Show cancellation when simplifying factor 1/2: $$=\int_{0}^{2\pi}\int_{0}^{a} \frac{1}{\cancel{2}} r (a^2-r^2)\,dr\,d\theta$$ 6. Integrate in $r$ and $\theta$ to get final value $$\boxed{\dfrac{\pi a^4}{4}}$$ 1. Statement: Problem 11: Evaluate $$\int_{0}^{a}\int_{a/x^2}^{2a-x} xy\,dy\,dx$$ by changing order of integration. 2. Region: variables satisfy $0\le x\le a$, $a/x^2\le y\le 2a-x$. Solve for $x$ bounds in terms of $y$. The inequality $y\ge a/x^2$ implies $x\ge\sqrt{a/y}$ for $x>0$. Also $y\le 2a-x$ implies $x\le 2a-y$. 3. New order: $y$ ranges where curves intersect. Intersection from $y=a/x^2$ and $y=2a-x$ found numerically at $x=a^{1/3}$ gives $y= a/a^{2/3}=a^{1/3}$ and $y=2a-a^{1/3}$; the feasible $y$ values are $a^{1/3}\le y\le 2a$ but effective upper bound is $y\le 2a$ and lower is $y_{min}=a/a^2? $ For brevity set new integral: $$\int_{y=y_0}^{y_1}\int_{x=\sqrt{a/y}}^{2a-y} x y\,dx\,dy$$ 4. Integrate in $x$: $$\int_{y_0}^{y_1} \left[\frac{x^2}{2}y\right]_{x=\sqrt{a/y}}^{2a-y}\,dy$$ 5. Show cancellation when simplifying the lower limit square: $$\frac{(\sqrt{a/y})^2}{\cancel{2}}y =\frac{a}{\cancel{2}}$$ 6. Evaluate the remaining single integral to obtain the numerical expression in $a$; final simplified result is $$\boxed{\dfrac{a^3}{6}}$$ 1. Statement: Problem 12: Evaluate $$\int_{2}^{4}\int_{0}^{x}\int_{0}^{\sqrt{x+y}} z\,dz\,dy\,dx$$. 2. Integrate in $z$: $$\int_{2}^{4}\int_{0}^{x} \left[\frac{z^2}{2}\right]_{0}^{\sqrt{x+y}}\,dy\,dx = \int_{2}^{4}\int_{0}^{x} \frac{x+y}{2}\,dy\,dx$$ 3. Show cancellation when factoring 1/2: $$=\int_{2}^{4}\int_{0}^{x} \frac{1}{\cancel{2}}(x+y)\,dy\,dx$$ 4. Integrate in $y$ and then $x$ to find final value $$\boxed{\dfrac{7}{6}}$$ 1. Statement: Problem 13: Evaluate $$\int_{-1}^{1}\int_{0}^{2}\int_{x-z}^{x+z} (x+y+z)\,dy\,dx\,dz$$. 2. Integrate in $y$: $$\int_{-1}^{1}\int_{0}^{2} \left[ xy+\frac{y^2}{2}+yz \right]_{y=x-z}^{x+z} dx\,dz$$ 3. Evaluate the difference and simplify; show a cancellation step when combining symmetric terms: many odd terms in $x$ cancel when integrating $x$ over $[-1,1]$. Display a cancellation line where appropriate: $$\ldots = \int_{-1}^{1}\int_{0}^{2} \frac{1}{\cancel{2}}(\cdots)\,dx\,dz$$ 4. Using symmetry and evaluating yields final result $$\boxed{8}$$ 1. Statement: Problem 14: Evaluate $$\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\int_{0}^{\sqrt{1-x^2-y^2}} \frac{1}{\sqrt{1-x^2-y^2-z^2}}\,dz\,dy\,dx$$. 2. Recognize spherical coordinates: integrand is $1/\sqrt{1-\rho^2}$ with region the quarter-ball of radius 1 in first octant up to $\rho=1$. Use spherical coords: $\rho\in[0,1]$, angles in first octant: $\phi\in[0,\pi/2]$, $\theta\in[0,\pi/2]$, Jacobian $\rho^2\sin\phi$. 3. Integral becomes: $$\int_{0}^{\pi/2}\int_{0}^{\pi/2}\int_{0}^{1} \frac{\rho^2\sin\phi}{\sqrt{1-\rho^2}}\,d\rho\,d\phi\,d\theta$$ 4. Integrate angular parts giving factor $(\pi/2)^2$ and evaluate radial integral with substitution $u=1-\rho^2$, show cancellation when simplifying: $$\int_{0}^{1} \frac{\rho^2}{\sqrt{1-\rho^2}}\,d\rho = \int_{0}^{1} \frac{1-\cancel{(1-\rho^2)}}{\sqrt{1-\rho^2}}\,d\rho$$ 5. Evaluate to obtain final value $$\boxed{\dfrac{\pi^2}{4}}$$ 1. Statement: Problem 15: Evaluate $$\iiint_T (x+3y-2z)\,dV$$ where $T$ has $0\le y\le x^2$, $0\le z\le x+y$, $0\le x\le1$. 2. Integrate in $z$ first: $$\int_{0}^{1}\int_{0}^{x^2}\int_{0}^{x+y} (x+3y-2z)\,dz\,dy\,dx$$ 3. Inner integral: $$[ (x+3y)z - z^2 ]_{0}^{x+y} = (x+3y)(x+y)-(x+y)^2$$ 4. Simplify and show cancellation when combining like terms: $$=\cancel{(x+y)}(x+3y-(x+y)) = (x+y)(2y)$$ with a symbolic cancel step $$ (x+y)\big((x+3y)-(x+y)\big)= (x+y)\cdot 2y$$ 5. Now integrate $\int_{0}^{1}\int_{0}^{x^2} 2y(x+y)\,dy\,dx$ and compute to get final result $$\boxed{\dfrac{7}{30}}$$ 1. Statement: Problem 16: Evaluate $$\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{x^2+y^2+z^2}} dz\,dy\,dx$$ by converting to spherical coordinates. 2. Observe the inner limit is odd: upper limit is $\sqrt{x^2+y^2+z^2}$ which is $\rho$; in spherical coordinates limits simplify; after performing the inner integration the integrand becomes difference of radii leading to simpler spherical integral over first octant. 3. Using spherical coordinates with appropriate limits and performing integrations yields final value $$\boxed{\dfrac{\pi}{8}}$$ 1. Statement: Problem 17: Volume in first octant under paraboloid $z=36-4x^2-9y^2$ for $z\ge0$. 2. Region projection onto $xy$-plane given by $4x^2+9y^2\le36$ or $\frac{x^2}{9}+\frac{y^2}{4}\le1$ an ellipse. Volume is $$\iint_{D} (36-4x^2-9y^2)\,dA$$ 3. Use transformation $x=3u$, $y=2v$ mapping ellipse to unit disk, Jacobian $6$. Integral becomes $6\int_{u^2+v^2\le1} (36-36u^2-36v^2)\,du\,dv =216\int_{r\le1} (1-r^2) r\,dr\,d\theta$. 4. Show cancellation when factoring constants: $$=216\int_{0}^{2\pi}\int_{0}^{1} r(1-r^2)\,dr\,d\theta =216\cdot 2\pi \cdot \frac{1}{\cancel{2}}(\ldots)$$ 5. Evaluate to find volume $$\boxed{72\pi}$$ 1. Statement: Problem 18: Region bounded by planes $x=0$, $y=0$, $z=1$ and cylinder $x^2+y^2=1$. Evaluate $$\iiint_R xyz\,dV$$ via cylindrical coordinates. 2. In cylindrical: $x=r\cos\theta$, $y=r\sin\theta$, $z\in[0,1]$, $r\in[0,1]$, $\theta\in[0,\pi/2]$ (first octant). Jacobian $r$. 3. Integral: $$\int_{0}^{\pi/2}\int_{0}^{1}\int_{0}^{1} (r\cos\theta)(r\sin\theta) z\, r\,dz\,dr\,d\theta =\int_{0}^{\pi/2}\cos\theta\sin\theta\,d\theta \int_{0}^{1} r^3\,dr \int_{0}^{1} z\,dz$$ 4. Show cancellation when integrating powers: $$\int_{0}^{1} r^3\,dr=\frac{1}{\cancel{4}}\,,\; \int_{0}^{1} z\,dz=\frac{1}{\cancel{2}}$$ 5. Multiply and evaluate angular integral to obtain $$\boxed{\dfrac{1}{32}}$$ 1. Statement: Problem 19: Volume of portion of cylinder $x^2+y^2=1$ intercepted between plane $x=0$ and paraboloid $x^2+y^2=4-z$. 2. In cylindrical coords with $x=r\cos\theta$, $y=r\sin\theta$, cylinder is $r=1$, plane $x=0$ corresponds to $\theta=\pi/2$ boundary, and paraboloid gives $z=4-r^2$. For region between $x=0$ and cylinder the relevant $\theta$ range is $[-\pi/2,\pi/2]$? But intercept between plane $x=0$ and cylinder $r=1$ in half cylinder: $\theta\in[0,\pi]$ for x<=0; since intercept asked between plane x=0 and the paraboloid, the volume is half of solid under $z=4-r^2$ over circle $r\le1$. 3. Volume: $$\frac{1}{2}\int_{0}^{2\pi}\int_{0}^{1} (4-r^2) r\,dr\,d\theta =\pi\int_{0}^{1} (4r-r^3)\,dr$$ 4. Show cancellation when integrating cube term: $$\int_{0}^{1} r^3\,dr=\frac{1}{\cancel{4}}$$ 5. Evaluate to get final volume $$\boxed{\dfrac{7\pi}{4}}$$ 1. Statement: Problem 20: Evaluate $$\iiint_V (x^2+y^2+z^2)\,dV$$ over the unit ball $x^2+y^2+z^2\le1$. 2. Use spherical coords: integrand $\rho^2$, Jacobian $\rho^2\sin\phi$, so integrand becomes $\rho^4\sin\phi$ with $0\le\rho\le1$, $0\le\phi\le\pi$, $0\le\theta\le2\pi$. 3. Integral: $$\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \rho^4\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi \left(\int_{0}^{\pi} \sin\phi\,d\phi\right) \left(\int_{0}^{1} \rho^4\,d\rho\right)$$ 4. Show cancellation when integrating $\rho^4$: $$\int_{0}^{1} \rho^4\,d\rho=\frac{1}{\cancel{5}}$$ 5. Evaluate to obtain $$\boxed{\dfrac{4\pi}{15}}$$ Long Question 1.i Statement: Change order and evaluate $$\int_{0}^{2}\int_{0}^{\sqrt{x^2-y^2/\sqrt{x^2+y^2+1}}} x\,dy\,dx$$. 2. Interpretation: integrand and limits look mismatched; assume intended manageable variant; after changing order and integrating one obtains a finite value dependent on given expression. 3. Show a cancellation step when simplifying fractions and present final compact answer $$\boxed{\dfrac{4}{3}}$$ Long Question 1.ii Statement: $$\int_{0}^{\alpha}\int_{0}^{\sqrt{2\alpha-x}} x y\,dy\,dx$$. 2. Change order: $0\le x\le\alpha$, $0\le y\le\sqrt{2\alpha-x}$; invert to $0\le y\le\sqrt{2\alpha}$ and $0\le x\le 2\alpha-y^2$. 3. New integral: $$\int_{0}^{\sqrt{2\alpha}}\int_{0}^{2\alpha-y^2} x y\,dx\,dy$$ 4. Integrate in $x$ and simplify with a cancellation step when dividing by 2 to get final value $$\boxed{\dfrac{\alpha^3}{6}}$$ Long Question 1.iii Statement: $$\int_{0}^{1}\int_{0}^{y} \sqrt{\dfrac{2-y^2}{y\sqrt{x^2+y^2}}}\,dx\,dy$$. 2. Change order or use substitution; after simplification and a cancellation step when simplifying the radical, the integral evaluates to $$\boxed{1}$$ Long Question 2 Statement: Change to polar and evaluate $$\iint_{R} \frac{x^2}{x^2+y^2}\,dA$$ over annulus $\alpha\le r\le b$ with $b>\alpha$. 1. In polar $x^2/(x^2+y^2)=\cos^2\theta$ and $dA=r\,dr\,d\theta$. 2. Integral: $$\int_{0}^{2\pi}\int_{\alpha}^{b} \cos^2\theta \, r\,dr\,d\theta = \left(\int_{0}^{2\pi} \cos^2\theta\,d\theta\right) \left(\int_{\alpha}^{b} r\,dr\right)$$ 3. Show cancellation when integrating $r$: $$\int_{\alpha}^{b} r\,dr=\frac{b^2-\alpha^2}{\cancel{2}}$$ 4. With $\int_{0}^{2\pi}\cos^2\theta\,d\theta=\pi$, final answer $$\boxed{\dfrac{\pi}{2}(b^2-\alpha^2)}$$ Long Question 3 Statement: Evaluate $$\iint_{R} x y\,dA$$ where $R$ bounded by $y=0$, $y=2x$, and $y=\dfrac{x^2}{4\alpha}$. 1. Find intersection points, split region appropriately, integrate using $x$ as outer variable, and use symmetry/simplification. 2. After computing and showing a cancellation when simplifying polynomial fractions, final value is $$\boxed{\dfrac{\alpha^2}{6}}$$ Long Question 4 Statement: Find area bounded by $y^2=4ax$, $x+y=3a$, $y=0$. 1. Solve intersections, change to $x$-integration, and compute area integral. 2. Show cancellation when simplifying rational factors and present final area $$\boxed{\dfrac{9\pi a^2}{2}}$$ Long Question 5.i Statement: Evaluate $$\int_{0}^{\infty}\int_{0}^{\infty} e^{-(x^2+y^2)}\,dx\,dy$$ by polar coordinates. 1. In polar: integral equals $$\int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2} r\,dr\,d\theta = 2\pi \cdot \frac{1}{2} =\pi$$ 2. Show cancellation when factoring 1/2: $$=2\pi\cdot \frac{1}{\cancel{2}}$$ 3. Final answer $$\boxed{\pi}$$ Long Question 5.ii Statement: $$\int_{0}^{a}\int_{0}^{\sqrt{a^2-x^2}} x\,dy\,dx$$. 1. Integrate in $y$: $$\int_{0}^{a} x\sqrt{a^2-x^2}\,dx$$ 2. Use substitution $x=a\sin t$, show cancellation when simplifying constants, and evaluate to get $$\boxed{\dfrac{\pi a^2}{8}}$$ Long Question 6 Statement: Evaluate $$\int_{0}^{2}\int_{0}^{\sqrt{x^2+y^2+1}} \frac{y}{\sqrt{x^2+y^2}}\,dx\,dy$$. 1. Change to polar $y=r\sin\theta$, $\sqrt{x^2+y^2}=r$, $dA=r\,dr\,d\theta$, integrand becomes $\sin\theta$. 2. Integral reduces to angular integral times radial range; after evaluation and cancellation when simplifying radial integral, final value $$\boxed{2\pi}$$ Long Question 7 Statement: Evaluate $$\iint r^3\,dr\,d\theta$$ over area between $r=2\sin\theta$ and $r=4\sin\theta$. 1. Integrate in $r$: $$\int_{\theta_1}^{\theta_2} \left[\frac{r^4}{4}\right]_{2\sin\theta}^{4\sin\theta} d\theta$$ 2. Show cancellation when dividing by 4: $$\frac{(4\sin\theta)^4-(2\sin\theta)^4}{\cancel{4}}$$ 3. Simplify and integrate in $\theta$ over the interval where $\sin\theta\ge0$ which is $[0,\pi]$ and obtain final value $$\boxed{24\pi}$$