Seja A a região que fica entre as curvas y= e^x e ln(3)y-2x=ln(3) a) fazer esboço da região b) verifique que as curvas se interseptam em dois pontos, um sobre o eixo y e outro sobr... Seja A a região que fica entre as curvas y= e^x e ln(3)y-2x=ln(3) a) fazer esboço da região b) verifique que as curvas se interseptam em dois pontos, um sobre o eixo y e outro sobre a reta x=ln(3). determine as coordenadas desses pontos c) determine as integrais que representam o volume do solido gerado pela rotação de A em torno do eixo y, utilizanddo o método dos discos e o metodo das cascas cilindricas. ( não é necessario resolver as integrais)
Understand the Problem
A questão descreve uma região A delimitada pelas curvas y = e^x e ln(3)y - 2x = ln(3). O problema pede para: a) Esboçar a região A. b) Verificar que as curvas se intersectam em dois pontos específicos: um sobre o eixo y e outro sobre a reta x = ln(3), e determinar as coordenadas desses pontos. c) Determinar as integrais que representam o volume do sólido gerado pela rotação de A em torno do eixo y, utilizando os métodos dos discos e das cascas cilíndricas (sem a necessidade de resolver as integrais).
Answer
a) Region bounded by $y=e^x$ and $y=\frac{2x}{\ln(3)} + 1$ b) $(0, 1)$ and $(\ln(3), 3)$ c) Disk method: $V = \pi \int_1^3 \left[ \left(\frac{\ln(3)(y - 1)}{2}\right)^2 - (\ln(y))^2 \right] dy$ Cylindrical shell method: $V = 2\pi \int_0^{\ln(3)} x \left(e^x - \frac{2x}{\ln(3)} - 1\right) dx$
Answer for screen readers
a) The region A is bounded by the curves $y=e^x$ and $y=\frac{2x}{\ln(3)} + 1$.
b) The intersection points are $(0, 1)$ and $(\ln(3), 3)$.
c) Disk method: $V = \pi \int_1^3 \left[ \left(\frac{\ln(3)(y - 1)}{2}\right)^2 - (\ln(y))^2 \right] dy$
Cylindrical shell method: $V = 2\pi \int_0^{\ln(3)} x \left(e^x - \frac{2x}{\ln(3)} - 1\right) dx$
Steps to Solve
- Sketch the region A
Sketch the graphs of $y = e^x$ and $\ln(3)y - 2x = \ln(3)$. The equation $\ln(3)y - 2x = \ln(3)$ can be rewritten as $y = \frac{2x + \ln(3)}{\ln(3)} = \frac{2}{\ln(3)}x + 1$. This is a linear equation. The graph of $y = e^x$ is a standard exponential curve.
- Find the intersection points
We need to find the $x$ values where $e^x = \frac{2x}{\ln(3)} + 1$.
It is given that they intersect when $x = 0$ and $x = \ln(3)$.
When $x = 0$, $y = e^0 = 1$, and $y = \frac{2(0)}{\ln(3)} + 1 = 1$. So, the point is $(0, 1)$. When $x = \ln(3)$, $y = e^{\ln(3)} = 3$, and $y = \frac{2\ln(3)}{\ln(3)} + 1 = 2 + 1 = 3$. So the point is $(\ln(3), 3)$.
- Set up integrals for the volume using the disk method
Since we are rotating around the y-axis, we need to express $x$ in terms of $y$. For $y = e^x$, we have $x = \ln(y)$. For $y = \frac{2x}{\ln(3)} + 1$, we have $y - 1 = \frac{2x}{\ln(3)}$, so $x = \frac{\ln(3)(y - 1)}{2}$.
The volume by the disk method is $$V = \pi \int_1^3 \left[ \left(\frac{\ln(3)(y - 1)}{2}\right)^2 - (\ln(y))^2 \right] dy$$
- Set up integrals for the volume using the cylindrical shell method
The height of the shell is the difference between the two curves, $e^x - \left(\frac{2x}{\ln(3)} + 1\right)$. The radius is $x$, and the thickness is $dx$.
The volume by the cylindrical shell method is $$V = 2\pi \int_0^{\ln(3)} x \left(e^x - \frac{2x}{\ln(3)} - 1\right) dx$$
a) The region A is bounded by the curves $y=e^x$ and $y=\frac{2x}{\ln(3)} + 1$.
b) The intersection points are $(0, 1)$ and $(\ln(3), 3)$.
c) Disk method: $V = \pi \int_1^3 \left[ \left(\frac{\ln(3)(y - 1)}{2}\right)^2 - (\ln(y))^2 \right] dy$
Cylindrical shell method: $V = 2\pi \int_0^{\ln(3)} x \left(e^x - \frac{2x}{\ln(3)} - 1\right) dx$
More Information
The disk method integrates along the y-axis, while the shell method integrates along the x-axis. Choosing the right method can simplify the integration process depending on the problem.
Tips
- Forgetting to square the radius in the disk method.
- Incorrectly solving for $x$ in terms of $y$ when using the disk method.
- Not setting up the limits of integration correctly.
- Incorrectly identifying the height of the cylindrical shell.
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