# Airshow4444's question at Yahoo! Answers regarding volumes by slicing

#### MarkFL

##### Administrator
Staff member
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.

#### MarkFL

##### Administrator
Staff member
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:

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