Volume of solids. Hello friends, today I’ll show how to use the method of triple integrals to find out the volume of solids. Have a look!!

Want to know more about multiple integrals? Check these out:

**Triple integrals | How to evaluate the triple integrals**

**Double integrals | How to evaluate the double integrals**

**Volume of solids**

Now here I’ll give some examples of that.

**Solved examples of the volume of solids**

Disclaimer: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

**Example 1**

According to Stroud and Booth (2013)*, “A solid is enclosed by the planes and the surface . Calculate the volume of the solid.”

**Solution**

Now here the solid is enclosed by the planes and the surface . And I have to calculate its volume. So I’ll start with the triple integral.

Thus the volume will be

First of all, I’ll integrate with respect to . So it will be

Next, I’ll substitute the limits. Now is the lower limit and is the upper limit. Therefore it will be

And that gives

Then I’ll integrate it with respect to to get

Now I’ll substitute the limits. Also is the lower limit and is the upper limit. So it will be

Then I’ll simplify it to get

Next, I’ll integrate it. So that means

Now I’ll substitute the limits. Also is the lower limit and is the upper limit. So it will be

Then I’ll simplify it to get

So that gives

Thus the volume of the given solid will be

Hence I can conclude that this is the solution to the given example.

Now I’ll give you another example.

**Example 2**

According to Stroud and Booth (2013)*, “A solid consists of vertical sides standing on the plane figure enclosed by and . The top is the surface . Find the volume of the solid so defined.”

**Solution**

Now in this example, the solid consists of vertical sides standing on the plane figure is enclosed by and . Also, the top is the surface . And I have to find out its volume. Again I’ll start with the triple integral.

Thus the volume will be

First of all, I’ll integrate with respect to . So it will be

Next, I’ll substitute the limits. Now is the lower limit and is the upper limit. Therefore it will be

And that gives

Then I’ll integrate it with respect to to get

Now I’ll substitute the limits. Also is the lower limit and is the upper limit. So it will be

Then I’ll simplify it to get

Next, I’ll integrate it. So that means

Now I’ll substitute the limits. Also is the lower limit and is the upper limit. So it will be

Then I’ll simplify it to get

Hence I can conclude that this is the solution to the given example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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