1. Evaluate $\displaystyle \int_{-1}^{1}{\int_{0}^{\sqrt{1-x^2}}{e^{-y^2-x^2}\;dy}\;dx}$

    $\displaystyle \pi\,\int_{0}^{1}{e^ {- r }\;dr}$ = $\displaystyle \left(1-e^ {- 1 }\right)\,\pi$ $\displaystyle {{\pi\,\int_{0}^{1}{r\,e^ {- r^2 }\;dr}}\over{2}}$ = $\displaystyle {{\left({{1}\over{2}}-{{e^ {- 1 }}\over{2}}\right)\,\pi}\over{2}}$ $\displaystyle \pi\,\int_{0}^{1}{r\,e^ {- r^2 }\;dr}$ = $\displaystyle \left({{1}\over{2}}-{{e^ {- 1 }}\over{2}}\right)\,\pi$ $\displaystyle {{\pi\,\int_{0}^{1}{e^ {- r }\;dr}}\over{2}}$ = $\displaystyle {{\left(1-e^ {- 1 }\right)\,\pi}\over{2}}$

  2. Evaluate $\int\!\!\!\int_R \left(y+x\right)\,e^{x-y} dxdy$ over $R = \{( v+u, u-v): 0 \le u\le 1, 0 \le v\le 1 \}$.

    $\displaystyle \left(2\,e-2\right)\,\int_{0}^{1}{e^{v}\;dv}$ = (e − 1)(2e − 2) $\displaystyle \int_{0}^{1}{e^{2\,v}\;dv}$ = $\displaystyle {{e^2}\over{2}}-{{1}\over{2}}$ $\displaystyle \left(e-1\right)\,\int_{0}^{1}{e^{v}\;dv}$ = $\displaystyle \left(e-1\right)^2$ $\displaystyle 2\,\int_{0}^{1}{e^{2\,v}\;dv}$ = $\displaystyle 2\,\left({{e^2}\over{2}}-{{1}\over{2}}\right)$

  3. Evaluate $\int\!\!\!\int\!\!\!\int_Rz dxdydz$ over the region $R$ bounded by the planes $x=0$,$y=0$,$z=0$, and 6z + 3y + 2x = 6.

    $\displaystyle \int_{0}^{3}{\int_{0}^{3-3\,x}{\int_{0}^{-{{y}\over{3}}-x+1}{z\;dz} \;dy}\;dx}$ = $\displaystyle -{{15}\over{8}}$

    $\displaystyle \int_{0}^{1}{\int_{0}^{2-{{2\,x}\over{3}}}{\int_{0}^{-{{y}\over{3}} -x+1}{z\;dz}\;dy}\;dx}$ = $\displaystyle {{377}\over{2916}}$

    $\displaystyle \int_{0}^{1}{\int_{0}^{3-3\,x}{\int_{0}^{-{{y}\over{2}}-{{x}\over{3 }}+1}{z\;dz}\;dy}\;dx}$ = $\displaystyle {{55}\over{288}}$

    $\displaystyle \int_{0}^{3}{\int_{0}^{2-{{2\,x}\over{3}}}{\int_{0}^{-{{y}\over{2}} -{{x}\over{3}}+1}{z\;dz}\;dy}\;dx}$ = $\displaystyle {{1}\over{4}}$

  4. Find the surface area for $z = x\,y+1$ above $R = \{(r\cos\theta,r\sin\theta): 0\le r\le 1, 0 \le\theta\le 2\,\pi \}$.

    $\displaystyle 2\,\pi\,\int_{0}^{1}{r\,\sqrt{r^2+1}\;dr}$ = $\displaystyle 2\,\left({{2^{{{3}\over{2}}}}\over{3}}-{{1}\over{3}}\right)\,\pi$ $\displaystyle 2\,\pi\,\int_{0}^{1}{r\,\sqrt{4\,r^2+1}\;dr}$ = $\displaystyle 2\,\left({{5^{{{3}\over{2}}}}\over{12}}-{{1}\over{12}}\right)\,\pi$ $\displaystyle 2\,\pi\,\int_{0}^{1}{r^3\;dr}$ = $\displaystyle {{\pi}\over{2}}$ $\displaystyle 2\,\pi\,\int_{0}^{1}{r\;dr}$ = $\displaystyle \pi$

  5. Find the equivalent expression of the region for $R = \{(x,y): -1 \le x\le 1, 0 \le y\le \sqrt{1-x^2} \}$.

    $R = \{(r\cos\theta,r\sin\theta): 0\le r\le 1, \pi \le\theta\le 2\,\pi \}$. $R = \{(r\cos\theta,r\sin\theta): 0\le r\le 1, 0 \le\theta\le \pi \}$. $R = \{(r\cos\theta,r\sin\theta): 0\le r\le 1, 0 \le\theta\le {{\pi}\over{2}} \}$.

  6. Find the surface area for $z = 2\,y+x^2$ above $R = \{(x,y): 0 \le x\le 1, 0 \le y\le x \}$.

    $\displaystyle 2\,\int_{0}^{1}{x^2\;dx}$ = $\displaystyle {{2}\over{3}}$ $\displaystyle \int_{0}^{1}{x^3+x^2\;dx}$ = $\displaystyle {{7}\over{12}}$ $\displaystyle \int_{0}^{1}{x\,\sqrt{4\,x^2+5}\;dx}$ = $\displaystyle -{{5^{{{3}\over{2}}}-27}\over{12}}$ $\displaystyle \sqrt{6}\,\int_{0}^{1}{x\;dx}$ = $\displaystyle {{\sqrt{6}}\over{2}}$

  7. Evaluate $\int\!\!\!\int_R y dxdy$ over $R = \{( {{u}\over{v}}, v): 1 \le u\le 4, \sqrt{u} \le v\le 2\,\sqrt{u} \}$.

    $\displaystyle \int_{1}^{4}{u^{{{3}\over{2}}}\;du}$ = $\displaystyle {{62}\over{5}}$ $\displaystyle {{3\,\int_{1}^{4}{u\;du}}\over{2}}$ = $\displaystyle {{45}\over{4}}$ $\displaystyle -{{\int_{1}^{4}{u\,\ln u-2\,\ln \left(2\,\sqrt{u}\right)\,u\;du} }\over{2}}$ = $\displaystyle {{15\,\ln 2}\over{2}}$ $\displaystyle \int_{1}^{4}{\sqrt{u}\;du}$ = $\displaystyle {{14}\over{3}}$

  8. Evaluate $\int\!\!\!\int_R x\,y dxdy$ over $R = \{(r\cos\theta,r\sin\theta): 0\le r\le 1, 0 \le\theta\le {{\pi}\over{2}} \}$.

    $\displaystyle {{\int_{0}^{1}{r^5\;dr}}\over{4}}$ = $\displaystyle {{1}\over{24}}$ $\displaystyle {{\int_{0}^{1}{r^3\;dr}}\over{2}}$ = $\displaystyle {{1}\over{8}}$ $\displaystyle {{\int_{0}^{1}{r^2\;dr}}\over{2}}$ = $\displaystyle {{1}\over{6}}$ $\displaystyle {{\int_{0}^{1}{r^4\;dr}}\over{4}}$ = $\displaystyle {{1}\over{20}}$

  9. Find the region $R$ bounded by the planes $x=0$,$y=0$,$z=0$, and 4z + 2y + 2x = 4.

    $R = \{(x,y,z): 0 \le x\le 2, 0 \le y\le 2-x, 0 \le z\le -{{y}\over{2}}-{{x}\over{2}}+1 \}$. $R = \{(x,y,z): 0 \le x\le 2, 0 \le y\le 3-x, 0 \le z\le -{{y}\over{3}}-{{x}\over{3}}+1 \}$. $R = \{(x,y,z): 0 \le x\le 3, 0 \le y\le 3-x, 0 \le z\le -{{y}\over{2}}-{{x}\over{2}}+1 \}$. $R = \{(x,y,z): 0 \le x\le 3, 0 \le y\le 2-x, 0 \le z\le -{{y}\over{3}}-{{x}\over{3}}+1 \}$.

  10. Evaluate $\int\!\!\!\int_R \left(2\,x-y\right)\,e^{2\,y+x} dxdy$ over $R = \{(r\cos\theta,r\sin\theta): 0\le r\le 1, \pi \le\theta\le {{3\,\pi}\over{2}} \}$.

    $\displaystyle -\int_{0}^{1}{e^ {- 2\,r }\,\left(e^{r}-1\right)\;dr}$ = $\displaystyle {{e^ {- 2 }\,\left(2\,e-1\right)}\over{2}}-{{1}\over{2}}$ $\displaystyle -\int_{0}^{1}{e^ {- 2\,r }\,\left(r\,e^{r}-r\right)\;dr}$ = $\displaystyle {{e^ {- 2 }\,\left(8\,e-3\right)}\over{4}}-{{3}\over{4}}$ $\displaystyle -\int_{0}^{1}{r\,e^ {- r }\,\left(e^{r}-1\right)\;dr}$ = $\displaystyle 1-{{e^ {- 1 }\,\left(e+4\right)}\over{2}}$ $\displaystyle -\int_{0}^{1}{e^ {- r }\,\left(e^{r}-1\right)\;dr}$ = $\displaystyle 1-e^ {- 1 }\,\left(e+1\right)$


Department of Mathematics
Last modified: 2025-10-23