Hi,
I have a problem i wasn't able to solve, i don't know how to calculate the volume of the generated body, when it is rotated around both the x and the y axis.
$\displaystyle A = \{(x,y):\sqrt{x}e^{x^2}~\le y \le 3,~ 0 \le x \le 1 \}$
Jones
Hi,
I have a problem i wasn't able to solve, i don't know how to calculate the volume of the generated body, when it is rotated around both the x and the y axis.
$\displaystyle A = \{(x,y):\sqrt{x}e^{x^2}~\le y \le 3,~ 0 \le x \le 1 \}$
Jones
I can show you how to calculate the volume when the curve rotates about the x-axis.
The general formula for a rotation volume with the x-axis as axis of rotation is:
$\displaystyle V_{rotx} = \pi\int_a^b y^2 dx$
Plug in the terms you know:
$\displaystyle V_{rotx}=\pi \int_0^1\left(\sqrt{x} \cdot e^{x^2} \right)^2 dx$
$\displaystyle V_{rotx}=\pi \int_0^1\left(x \cdot e^{2x^2} \right) dx$
Now use integration by substitution:
$\displaystyle u = 2x^2~\implies~\frac{du}{dx} = 4x~\implies~du = 4x dx$
The integral becomes:
$\displaystyle V_{rotx}=\frac{\pi}{4} \int_0^1\left(4x \cdot e^{2x^2} \right) dx$
$\displaystyle V_{rotx}=\frac{\pi}{4} \int_0^1\left( e^{u} \right) du$
Can you take it from here?
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There is missing at least one value to calculate the rotation volume if the y-axis is the axis of rotation.
I've had some difficulties to understand this line:
$\displaystyle A = \{(x,y):\sqrt{x}e^{x^2}~\le y \le 3,~ 0 \le x \le 1 \}$
If you mean
$\displaystyle y = f(x)= \sqrt{x} \cdot e^{x^2}\ ,\ 0\leq x \leq 1$
then the property $\displaystyle y \leq 3$ is unnecessary because f(1) = e < 3.
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To calculate the rotation volume with the y-axis as axis of rotation you have to use the general equation:
$\displaystyle V_{roty} = \int_a^b \pi \cdot x^2 dy$
Since $\displaystyle \frac{dy}{dx} = y'~\implies~dy = y' \cdot dx$
this equation becomes:
$\displaystyle V_{roty} = \int_a^b \pi \cdot x^2 y' \cdot dx$
But unfortunately the integrand becomes very ugly so I can only provide you with an approximative result: $\displaystyle V_{roty} \approx 3.994$