Parameterization & Surface Integrals - Multivariable Calculus

Card 0 of 8

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

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We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$