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{\bf HW 8\\Math 261, S19}

Please see the course syllabus for details on how to turn in your homework assignments.  This one is due at the beginning of your class on \underbar{\bf Friday,  April 12}.


\begin{enumerate}
\item (5 pts.) {\textsc{True or False:}}
\begin{enumerate}
\item Let $R$ denote a plane region, and $(u,v) = (u(x,y), v(x,y))$ be a different set of coordinates for the Cartesian plane. Then for any function $F(u,v)$ $$ \int_R F(u,v) du dv = \int_R F (u(x,y), v(x,y)) dxdy.$$
\item Let $R$ denote a plane region, and $(u,v) = (u(x,y), v(x,y))$ be a different set of coordinates for the Cartesian plane. Then $$ \int_R du dv = \int_R  |u_xv_y-u_y v_x| dxdy.$$
\item Let $R$ denote a square of sidelength $2$ defined by the inequalities $|x|\leq 1, |y|\leq 1$, and $(u,v) =(3y,2x)$. Then  the area of $R$ is computed as $$ \int_{-2}^{2}\int_{-3}^{3} du dv. $$
\item Let $R$ denote a square of sidelength $2$ defined by the inequalities $|x|\leq 1, |y|\leq 1$, and $(u,v) =(3y,2x)$. Then  the area of $R$ is computed as $$ \int_{-2}^{2}\int_{-3}^{3} (1/6) du dv. $$
\item Let $R$ denote a square of sidelength $2$ defined by the inequalities $|x|\leq 1, |y|\leq 1$, and $(u,v) =(3y,2x)$. Then  the area of $R$ is computed as $$ \int_{-1}^{1}\int_{-1}^1 (1/6) du dv. $$
\end{enumerate}

\item (3 pts.) Using {\em cylindrical} coordinates, set up the integral to find the volume of the region enclosed by the vertical cylinder $x^2+y^2=4$ and the planes $z=0$ and $y+z=4$.  Do {\bf NOT} evaluate the integral; just set it up.

\item (3 pts.) Using {\em spherical} coordinates, set up the integral to find the volume of the region enclosed by the vertical cylinder $x^2+y^2=4$ and the planes $z=0$ and $z=2$.  Do {\bf NOT} evaluate the integral; just set it up.  

\item (3 pts.) Use {\em cylindrical} coordinates and set up the integral to find the volume of the region enclosed by a circular cone of base radius $1$ and height $2$.  Do {\bf NOT} evaluate the integral; just set it up.  Your set-up should include a description of how the cone is positioned in $\mathbb{R}^3$.

\item (3 pts.) Consider using the substitution $\begin{cases} x=u-v,\\ y=2u+v\end{cases}$ for the integral of $x+y^2-2$.   What is the {\em integrand} in terms of $u$ and $v$?  (Don't bother with the integral signs, the bounds, or the $du\ dv$.)
  
  
\item (3 pts.) Using the same substitution as in the previous problem, suppose the $(x,y)$ region over which we wish to integrate includes the boundary line $2x - y = 3$.  Convert this line into a $(u,v)$ boundary line.





\end{enumerate}

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