Double Integrations on Plane - Multivariable Calculus

Card 1 of 20

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

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

using the transformation , and .

Tap to reveal answer

Answer

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

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

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

$4u=0\rightarrow u=0$

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

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

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

Next we need to calculate the Jacobian.

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

Now we can put the integral together.

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

$=\int_${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$

$=\int_${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$

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

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

using the transformation , and .

Tap to reveal answer

Answer

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

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

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

$4u=0\rightarrow u=0$

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

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

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

Next we need to calculate the Jacobian.

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

Now we can put the integral together.

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

$=\int_${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$

$=\int_${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$

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

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

using the transformation , and .

Tap to reveal answer

Answer

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

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

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

$4u=0\rightarrow u=0$

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

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

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

Next we need to calculate the Jacobian.

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

Now we can put the integral together.

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

$=\int_${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$

$=\int_${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$

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

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

using the transformation , and .

Tap to reveal answer

Answer

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

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

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

$4u=0\rightarrow u=0$

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

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

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

Next we need to calculate the Jacobian.

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

Now we can put the integral together.

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

$=\int_${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$

$=\int_${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$

← Didn't Know|Knew It →

Question

Convert the following into spherical coordinates.

$x=\sin(t)$

$y=\cos(t)$

Tap to reveal answer

Answer

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

Now lets apply this to our problem.

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

Question

Calculate the following Integral.

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx$

Tap to reveal answer

Answer

$\int_${0}^{1}$\int_${x}^{x^2}$ $xy^2$ dydx=\int_${0}^{1}$\Big(\int_${x}^{x^2}$ $xy^2$ dy\Big)dx$

Lets deal with the inner integral first.

$\int_${x}^{x^2}$$xy^2$ dy$

$\int_${x}^{x^2}$$xy^2$ $dy=\frac{1}{3}$y^3x$\Big|_${x}^{x^2}$$

Now we evaluate this expression in the outer integral.

$\int_${0}^{1}$ $$\frac{1}{3}$x^7$$-$\frac{1}{3}$x^4dx$$

$=\frac{1}{3}$\int_${0}^{1}$ $x^7$$-x^4dx$$

$=\frac{1}{3}$\Big( $$\frac{1}{8}$x^8$$-$\frac{1}{5}$x^5$\Big)\Big|_${0}^{1}$$

$=\frac{1}{3}$\Big($\frac{1}{8}$-$\frac{1}{5}$\Big)-$\frac{1}{3}$\Big(0-0\Big)$

$=\frac{1}{3}$\Big($\frac{5}{40}$-$\frac{8}{40}$\Big)$

$=\frac{1}{3}$\Big(-$\frac{3}{40}$\Big)=-$\frac{3}{120}$$

← Didn't Know|Knew It →

0

Knew It