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

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

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

    $\displaystyle \int_{1}^{4}{u^2\;du}$ = 21 $\displaystyle {{3\,\int_{1}^{4}{u^2\;du}}\over{2}}$ = $\displaystyle {{63}\over{2}}$ $\displaystyle \int_{1}^{4}{u\,\ln \left(2\,u\right)-u\,\ln u\;du}$ = $\displaystyle {{15\,\ln 2}\over{2}}$ $\displaystyle \int_{1}^{4}{u\;du}$ = $\displaystyle {{15}\over{2}}$

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

    $\displaystyle {{\int_{0}^{3}{r^4\;dr}}\over{4}}$ = $\displaystyle {{243}\over{20}}$ $\displaystyle -{{\int_{0}^{3}{r^4\;dr}}\over{3}}$ = $\displaystyle -{{81}\over{5}}$ $\displaystyle -{{\int_{0}^{3}{r^3\;dr}}\over{3}}$ = $\displaystyle -{{27}\over{4}}$ $\displaystyle {{\int_{0}^{3}{r^5\;dr}}\over{4}}$ = $\displaystyle {{243}\over{8}}$

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

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

  5. Find the region $R$ bounded by the planes $x=0$,$y=0$,$z=0$, and 6z + 2y + 3x = 6.

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

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

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

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

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

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

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

  9. Evaluate $\int\!\!\!\int\!\!\!\int_Rz dxdydz$ over the region $R$ bounded by the planes $x=0$,$y=0$,$z=0$, and 4z + 2y + 2x = 4.

    $\displaystyle \int_{0}^{2}{\int_{0}^{3-x}{\int_{0}^{-{{y}\over{3}}-{{x}\over{3}}+ 1}{z\;dz}\;dy}\;dx}$ = $\displaystyle {{10}\over{27}}$

    $\displaystyle \int_{0}^{3}{\int_{0}^{3-x}{\int_{0}^{-{{y}\over{2}}-{{x}\over{2}}+ 1}{z\;dz}\;dy}\;dx}$ = $\displaystyle {{9}\over{32}}$

    $\displaystyle \int_{0}^{2}{\int_{0}^{2-x}{\int_{0}^{-{{y}\over{2}}-{{x}\over{2}}+ 1}{z\;dz}\;dy}\;dx}$ = $\displaystyle {{1}\over{6}}$

    $\displaystyle \int_{0}^{3}{\int_{0}^{2-x}{\int_{0}^{-{{y}\over{3}}-{{x}\over{3}}+ 1}{z\;dz}\;dy}\;dx}$ = $\displaystyle {{23}\over{72}}$

  10. Find the Jacobian of the transformation $\displaystyle x={{u}\over{v}}$ and y = v

    −4 u $\displaystyle {{1}\over{v}}$ $\displaystyle u\,v^2+4\,u^2\,v$


Department of Mathematics
Last modified: 2026-10-05