Solution : The formula for the total suface area of cube of side a is \(6a^2\). And the formula for the lateral surface area of cube is \(4a^2\). Read Post : Formula for Surface Area of Cube – Derivation & Example Similar Questions What is the Formula for Volume of Cuboid ? What is the …
Solution : The formula for surface area of a cuboid of length l, breadth b and height h is 2(lb + bh + hl). This is the total surface area of cuboid. The lateral surface area of cuboid is 2(l + b)h. Read Post : Formula for Surface Area of Cuboid – Derivation & Example …
What is the Formula for Surface Area of a Cuboid ? Read More »
Solution : The formula for volume of a cube of side a is \(a^3\). Read Post : Formula for Volume of Cube – Derivation & Example Similar Questions What is the Formula for Volume of Cuboid ? What is the Formula for Surface Area of Cube ? What is the Formula for Surface Area of …
Solution : The formula for volume of cylinder of base radius r and height h is \(\pi r^2 h\). Read Post : Formula for Volume of Cylinder – Definition & Examples Similar Questions What is the Formula for Volume of Cuboid ? What is the Formula for Surface Area of Cube ? What is the …
Solution : The curved surface area of cylinder is \(2\pi rh\) And the total surface are of cylinder is \(2\pi r(h + r)\). Read Post : Formula for Surface Area of cylinder – Derivation & Example Similar Questions What is the Formula for Volume of Cuboid ? What is the Formula for Surface Area of …
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Solution : Let I = \(\int\) \(tan^{-1}\sqrt{x}\).1 dx By Applying integration by parts, Taking \(tan^{-1}\sqrt{x}\) as first function and 1 as second function. Then I = \(tan^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(tan^{-1}\sqrt{x}\) \(\int\) 1 dx } dx I = x\(tan^{-1}\sqrt{x}\) – \(\int\) \(1\over 2(1+x)\sqrt{x}\) . x dx Let \(\sqrt{x}\) = t \(1\over 2\sqrt{x}\) …
Solution : Let I = \(\int\) x\(tan^{-1}x\) dx By using Integration by parts rule, Taking tan inverse x as first function and x as second function. Then, I = (\(tan^{-1}x\)) \(\int\) x dx – \(\int\){\({d\over dx}\)(\(tan^{-1}x\) \(\int\) x dx} dx I = (\(tan^{-1}x\))\(x^2\over 2\) – \(\int\)\({1\over 1 + x^2}\) \(\times\) \(x^2\over 2\) dx \(\implies\) I …
Solution : We know that the vector equation of line passing through two points with position vectors \(\vec{a}\) and \(\vec{b}\) is, \(\vec{r}\) = \(\lambda\) \((\vec{b} – \vec{a})\) Here \(\vec{a}\) = \(3\hat{i} + 4\hat{j} – 7\hat{k}\) and \(\vec{b}\) = \(\hat{i} – \hat{j} + 6\hat{k}\). So, the vector equation of the required line is \(\vec{r}\) = (\(3\hat{i} …
Solution : We have, \(\vec{a}\) = \(4\hat{i} – 3\hat{j} + 5\hat{k}\) and \(\vec{b}\) = \(3\hat{i} + 4\hat{j} + 5\hat{k}\) Let \(\theta\) is the angle between the given vectors. Then, cos\(\theta\) = \(\vec{a}.\vec{b}\over |\vec{a}||\vec{b}|\) \(\implies\) cos\(\theta\) = \(12 – 12 + 25\over \sqrt{16 + 9 + 25} \sqrt{16 + 9 + 25}\) = \(1\over 2\) \(\implies\) …
Solution : We have \(\vec{a}\) = \(2\hat{i}+2\hat{j}-\hat{k}\) and \(\vec{b}\) = \(6\hat{i}-3\hat{j}+2\hat{k}\) \(\vec{a}\).\(\vec{b}\) = (\(2\hat{i}+2\hat{j}-\hat{k}\)).(\(6\hat{i}-3\hat{j}+2\hat{k}\)) = (2)(6) + (2)(-3) + (-1)(2) = 12 – 6 – 2 = 4 Similar Questions Find the angle between the vectors with the direction ratios proportional to 4, -3, 5 and 3, 4, 5. Find the vector equation of a …
