Dated: 30-10-2024

Examples

Example

Find the volume of the solid common to both cylinders \(x^2 + y^2 = 25\) and \(x^2 + z^2 = 25\)

Solution

The region \(R\) which we will be integrating¹ over, is bounded by \(x^2 + y^2 = 25\)
The surface which will limits our integral¹ is the other cylinder.

\[x^2 + z^2 = 25\]

\[z = \sqrt{25 - x^2} \]

This is our height limit, which is our surface.

\[\because x^2 + y^2 = 5^2\]

\[\text{radius } = 5\]

\[\because V = \iint_R f(x, y) dxdy\]

We will be integrating¹ with respect to \(y\) in the end.
Therefore, we need numeric bounds for the integral¹ so we can get a numeric volume.
Since the radius is \(5\), \(0 \le y \le 5\).
For the inner integral,¹ we need bounds to be a function² of \(y\).

\[\because x^2 +y^2 = 25\]

\[y = \sqrt{25 - x^2}\]

Plugging in the values, we get

\[V = \int_0^5\int_{0}^{\sqrt{25 - x^2}} (\sqrt{25 - x^2})dxdy\]

Evaluate this and we will get

\[= \frac {2000} 3\]

Area Calculated as `Double Integral`³

We know that

\[\text{volume} = \text{area of R} \times \text{height}\]

If we want to find the area of \(R\) then height is \(1\).

\[\iint_R dA = \text{area of R} \times 1\]

\[\text{area of R} = \iint_R dA\]

Example

Find the area bounded by parabola \(y = x^2\) and line⁴ \(y = x + 2\)

Solution

Trying to find the intersection points between both equations, we will get \(x = -1, 2\).

\[A = \int_{-1} ^ 2 \int_{x^2}^{x + 2} dy dx\]

\[= \int_{-1}^2 \left[y\right]_{x^2}^{x + 2} dx\]

\[=\int_{-1}^2 (x + 2 - x^2)dx\]

\[= \left[\frac{x^2} 2 + 2x - \frac{x^3} 3\right]_{-1}^2\]

\[= \frac 9 2\]

Examples

Example

Solution

Area Calculated as Double Integral3

Example

Solution

References

Area Calculated as `Double Integral`³