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Multivariable Calculus : Change of Variables

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Example Questions

Example Question #1 : Change Of Variables

Convert the following into spherical coordinates.

$x=\sin(t)$ $\displaystyle x=\sin(t)$

$y=\cos(t)$ $\displaystyle y=\cos(t)$

z=1 $\displaystyle z=1$

Possible Answers:

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$ $\displaystyle \phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

$\rho=1$ $\displaystyle \rho=1$

$\phi=\frac{\pi}{2}$ $\displaystyle \phi=\frac{\pi}{2}$

$\theta=\tan(\cot(t))$ $\displaystyle \theta=\tan(\cot(t))$

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$

$\phi=0$ $\displaystyle \phi=0$

$\theta=\tan(\cot(t))$ $\displaystyle \theta=\tan(\cot(t))$

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$ $\displaystyle \phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

$\rho=2$ $\displaystyle \rho=2$

$\phi=-\frac{\pi}{4}$ $\displaystyle \phi=-\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

Correct answer:

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$

$\phi=\frac{\pi}{4}$ $\displaystyle \phi=\frac{\pi}{4}$

$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

Explanation:

In order to convert to spherical coordinates , we need to remember the conversion equations.

$\rho=\sqrt{x^2+y^2+z^2}$ $\displaystyle \rho=\sqrt{x^2+y^2+z^2}$

$\theta=\tan^{-1}\Big(\frac{y}{x}\Big)$ $\displaystyle \theta=\tan^{-1}\Big(\frac{y}{x}\Big)$

$\phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$ $\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$

Now lets apply this to our problem.

$\rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$ $\displaystyle \rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$\phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}$ $\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}$

$\theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$

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Example Question #2 : Change Of Variables

Evaluate $\int \int_R 3x+2y \ dA$ $\displaystyle \int \int_R 3x+2y \ dA$, where $\displaystyle R$ is the trapezoidal region with vertices given by (0,0) $\displaystyle (0,0)$, (5,0) $\displaystyle (5,0)$, $(\frac{5}{2}, \frac{5}{2})$ $\displaystyle (\frac{5}{2}, \frac{5}{2})$, and $(\frac{5}{2}, -\frac{5}{2})$ $\displaystyle (\frac{5}{2}, -\frac{5}{2})$,

using the transformation x=2u+3v $\displaystyle x=2u+3v$, and y=2u-3v $\displaystyle y=2u-3v$.

Possible Answers:

$\frac{375}{2}$ $\displaystyle \frac{375}{2}$

$\frac{375}{4}$ $\displaystyle \frac{375}{4}$

$\frac{35}{4}$ $\displaystyle \frac{35}{4}$

$\displaystyle 0$

$\frac{37}{4}$ $\displaystyle \frac{37}{4}$

Correct answer:

$\frac{375}{4}$ $\displaystyle \frac{375}{4}$

Explanation:

The first thing we have to do is figure out the general equations for the lines that create the trapezoid.

y=x $\displaystyle y=x$

y=-x $\displaystyle y=-x$

y=-x+5 $\displaystyle y=-x+5$

y=x-5 $\displaystyle y=x-5$

Now we have the general equations for out trapezoid, now we need to plug in our transformations into these equations.

y=x $\displaystyle y=x$

$\displaystyle 2u+3v=2u-3v$

$-6v=0\rightarrow v=0$ $\displaystyle -6v=0\rightarrow v=0$

y=-x $\displaystyle y=-x$

$\displaystyle 2u+3v=-2u+3v$

$4u=0\rightarrow u=0$ $\displaystyle 4u=0\rightarrow u=0$

y=-x+5 $\displaystyle y=-x+5$

$\displaystyle 2u+3v=-2u+3v+5$

$4u=5\rightarrow u=\frac{5}{4}$ $\displaystyle 4u=5\rightarrow u=\frac{5}{4}$

y=x-5 $\displaystyle y=x-5$

$\displaystyle 2u+3v=2u-3v-5$

$-6v=-5\rightarrow v=\frac{5}{6}$ $\displaystyle -6v=-5\rightarrow v=\frac{5}{6}$

So our region is a rectangle given by $0 \leq u \leq \frac{5}{4}$ $\displaystyle 0 \leq u \leq \frac{5}{4}$, $0 \leq v \leq \frac{5}{6}$ $\displaystyle 0 \leq v \leq \frac{5}{6}$

Next we need to calculate the Jacobian.

$\begin{vmatrix} 2 & 3\\ 2 & -3 \end{vmatrix} =-6-6=-12$ $\displaystyle \begin{vmatrix} 2 & 3\\ 2 & -3 \end{vmatrix} =-6-6=-12$

Now we can put the integral together.

$\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv$ $\displaystyle \int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv$

$=\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv$ $\displaystyle =\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv$

$=\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv$ $\displaystyle =\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv$

$=\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv$ $\displaystyle =\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv$

$=\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv$ $\displaystyle =\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv$

$=\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv$ $\displaystyle =\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv$

$=\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}$ $\displaystyle =\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}$

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Example Question #1 : Double Integrations On Plane

Convert the following into spherical coordinates.

$x=\sin(t)$ $\displaystyle x=\sin(t)$

$y=\cos(t)$ $\displaystyle y=\cos(t)$

z=1 $\displaystyle z=1$

Possible Answers:

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$
$\phi=\frac{\pi}{4}$ $\displaystyle \phi=\frac{\pi}{4}$
$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

$\rho=2$ $\displaystyle \rho=2$
$\phi=-\frac{\pi}{4}$ $\displaystyle \phi=-\frac{\pi}{4}$
$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

$\rho=1$ $\displaystyle \rho=1$
$\phi=\frac{\pi}{2}$ $\displaystyle \phi=\frac{\pi}{2}$
$\theta=\tan(\cot(t))$ $\displaystyle \theta=\tan(\cot(t))$

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$
$\phi=0$ $\displaystyle \phi=0$
$\theta=\tan(\cot(t))$ $\displaystyle \theta=\tan(\cot(t))$

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$
$\phi=-\frac{\pi}{4}$ $\displaystyle \phi=-\frac{\pi}{4}$
$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

Correct answer:

$\rho=\sqrt{2}$ $\displaystyle \rho=\sqrt{2}$
$\phi=\frac{\pi}{4}$ $\displaystyle \phi=\frac{\pi}{4}$
$\theta=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}(\cot(t))$

Explanation:

In order to convert to spherical coordinates , we need to remember the conversion equations.

$\rho=\sqrt{x^2+y^2+z^2}$ $\displaystyle \rho=\sqrt{x^2+y^2+z^2}$

$\theta=\tan^{-1}\Big(\frac{y}{x}\Big)$ $\displaystyle \theta=\tan^{-1}\Big(\frac{y}{x}\Big)$

$\phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$ $\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)$

Now lets apply this to our problem.

$\rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$ $\displaystyle \rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}$

$\theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$ $\displaystyle \theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))$

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Multivariable Calculus : Change of Variables

Study concepts, example questions & explanations for Multivariable Calculus

All Multivariable Calculus Resources

Example Questions

Example Question #1 : Change Of Variables

Example Question #2 : Change Of Variables

Example Question #1 : Double Integrations On Plane

All Multivariable Calculus Resources

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