Parameterization & Surface Integrals - Multivariable Calculus

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Question

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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Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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}$)=0$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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}$)=0$

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