Airshow4444's question at Yahoo Answers regarding volumes by slicing

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In summary, to find the volume of a solid using calculus, we can use the formula V=\int_{a}^{b}A(x)\,dx, where A(x) represents the area of a cross-section of the solid and a and b are the lower and upper limits of integration. In the given problem, the base of the solid is an elliptical region with boundary curve 25x^2+36y^2=900 and the cross-sections perpendicular to the x-axis are isosceles right triangles. By rewriting the base in standard form and using the given information, we can determine the base, height, and area of the triangular cross-section. Finally, by
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Here is the question:

Volume of a Solid using Calculus?


I am quite confused as to how to approach this. I know the volume will be the area between the curves but I haven't seen a problem like this before... Any help?

The base of a certain solid is an elliptical region with boundary curve 25x2+36y2=900. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

Use the formula V=∫baA(x)dx to find the volume of the solid.

The lower limit of integration is a =
The upper limit of integration is b =
The base of the triangular cross-section is the following function of x:
The height of the triangular cross-section is the following function of x:
The area of the triangular cross-section is A(x)=
Thus the volume of the solid is V=

I have posted a link there to this topic so the OP can see my work.
 
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Hello Airshow4444,

I would first rewrite the given elliptical base in standard form so we may determine the length of the horizontal axis:

\(\displaystyle 25x^2+36y^2=900\)

Divide through by $900$:

\(\displaystyle \frac{x^2}{6^2}+\frac{y^2}{5^2}=1\)

Hence, we see the lower limit of integration is $a=-6$ and $b=6$.

The base $B$ of the triangular cross-section for a given $x$ is:

\(\displaystyle B(x)=2y(x)=2\frac{5}{6}\sqrt{6^2-x^2}=\frac{5}{3}\sqrt{6^2-x^2}\)

The height $h$ of the triangular cross-section for a given $x$ is:

\(\displaystyle h(x)=y(x)=\frac{5}{6}\sqrt{6^2-x^2}\)

The area $A$ of the triangular cross-section for a given $x$ is:

\(\displaystyle A(x)=\frac{1}{2}B(x)h(x)=\frac{25}{36}\left(6^2-x^2 \right)\)

Hence, the volume of the solid is given by:

\(\displaystyle V=\frac{25}{36}\int_{-6}^{6}6^2-x^2\,dx=\frac{25}{16}\int_{0}^{6}6^2-x^2\,dx\)

Note: the integrand is an even function, therefore we may apply the even function rule.

Applying the FTOC, we find:

\(\displaystyle V=\frac{25}{18}\left[6^2x-\frac{1}{3}x^3 \right]_0^6=\frac{25}{18}\cdot6^3\left(1-\frac{1}{3} \right)=200\)
 

FAQ: Airshow4444's question at Yahoo Answers regarding volumes by slicing

What does "volumes by slicing" mean?

"Volumes by slicing" refers to a mathematical concept where the volume of a three-dimensional object is calculated by slicing it into multiple two-dimensional sections and summing their areas.

How is the volume of an object calculated using slicing?

The volume of an object can be calculated using slicing by multiplying the area of each two-dimensional section by its thickness and then adding all of the resulting volumes together. This method is often used for irregularly shaped objects that cannot be easily measured using traditional volume formulas.

What are the benefits of using slicing to calculate volume?

Slicing allows for the calculation of the volume of irregularly shaped objects that cannot be easily measured using traditional methods. It also provides a more accurate representation of the object's volume compared to estimates or approximations based on its shape.

Can slicing be used to calculate the volume of any object?

Slicing can be used to calculate the volume of most objects, but the accuracy of the result may vary depending on the complexity and irregularity of the object's shape. In some cases, other methods such as integration may be more appropriate for calculating volume.

Are there any limitations to using slicing for volume calculations?

One limitation of using slicing for volume calculations is that it requires the object to be divided into slices, which may not always be possible or practical. Additionally, the accuracy of the result may be affected by the thickness of the slices used, so smaller slices may be necessary for more precise calculations.

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