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Multivariable Calculus : Parameterization & Surface Integrals

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Example Question #1 : Triple Integration Of Surface

Evaluate $\int \int \int_D y \ dV$ $\displaystyle \int \int \int_D y \ dV$, where $\displaystyle D$ is the region below the plane z=x+1 $\displaystyle z=x+1$ , above the $\displaystyle xy$ plane and between the cylinders $\displaystyle x^2+y^2=1$, and $\displaystyle x^2+y^2=9$.

Possible Answers:

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

$\pi$ $\displaystyle \pi$

$2 \pi$ $\displaystyle 2 \pi$

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

$\displaystyle 0$

Correct answer:

$\displaystyle 0$

Explanation:

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the $\displaystyle xy$ plane, this means we are above z=0 $\displaystyle z=0$.

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

The region $\displaystyle D$ is between two circles $\displaystyle x^2+y^2=1$, and $\displaystyle x^2+y^2=9$.

This means that

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

$1\leq r \leq 3$ $\displaystyle 1\leq r \leq 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

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Example Question #1 : Parameterization & Surface Integrals

Possible Answers:

$\displaystyle 0$

$2 \pi$ $\displaystyle 2 \pi$

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

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

$\pi$ $\displaystyle \pi$

Correct answer:

$\displaystyle 0$

Explanation:

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the $\displaystyle xy$ plane, this means we are above z=0 $\displaystyle z=0$.

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

The region $\displaystyle D$ is between two circles $\displaystyle x^2+y^2=1$, and $\displaystyle x^2+y^2=9$.

This means that

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

$1\leq r \leq 3$ $\displaystyle 1\leq r \leq 3$

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

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

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

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

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

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

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

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Multivariable Calculus : Parameterization & Surface Integrals

Study concepts, example questions & explanations for Multivariable Calculus

All Multivariable Calculus Resources

Example Questions

Example Question #1 : Triple Integration Of Surface

Example Question #1 : Parameterization & Surface Integrals

All Multivariable Calculus Resources

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