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ineedhelpnow
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if I am finding the surface area, and I am given the equation in term of x=g(y) about the x-axis, do i have to solve for y or can i just integrate in terms of y? would i just use dx/dy instead?
ineedhelpnow said:that's how i did the problem:
$x=\frac{1}{3} (y^2+2)^{3/2}$ $1\le y \le 3$ about the x axis
$x'= \frac{(y^2+2)^{3/2}}{3} = \frac{ \frac{3(y^2+2)^{1/2}}{2}}{3}dy(y^2+2 ) = \frac{(y^2+2)^{1/2}*2y}{2} = y \sqrt{y^2+2}$
$S=\int_{1}^{3} \ 2\pi y \sqrt{1+y^2(y^2+2)}dy=2\pi \int_{1}^{3} \ y\sqrt{1+y^4+2y^2}dy=2\pi\int_{1}^{3} \ y\sqrt{(y^2+1)^2}dy=2\pi \int_{1}^{3} \ y(y^2+1)dy=2\pi \int_{1}^{3} \ (y^3+y) dy=2\pi [\frac{y^4}{4}+\frac{y^2}{2}]_{1}^{3} = 48 \pi$
is it correct?
In order to find the surface area from an equation in terms of x=g(y), you will need to use the formula SA = 2π∫g(y)√(1+(g'(y))^2)dy. This formula involves integrating the equation g(y) and its derivative g'(y) in order to find the surface area.
The equation x=g(y) represents the cross-sectional area of the solid figure at a specific value of y. This equation is used to determine the surface area of a solid figure that varies along the y-axis.
Finding the surface area from an equation in terms of x=g(y) is important because it allows you to calculate the surface area of complex shapes that cannot be easily measured or approximated. This method is especially useful in fields such as engineering, architecture, and physics.
Yes, this method can be used to find the surface area of any solid figure as long as the cross-sectional area can be expressed as an equation in terms of x=g(y). However, the integral required to calculate the surface area may be difficult or impossible to solve for some complex figures.
One limitation of this method is that it can only be used for figures that vary along the y-axis. It cannot be used for figures that vary along the x-axis or in multiple directions. Additionally, as mentioned before, the integral required to calculate the surface area may be difficult or impossible to solve for some complex shapes.